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At each point $x\in M$ the differential $df_x: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $df_x(T_xM)$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

One way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, if we can take $U = \Bbb R^2$ and defined $r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)

At each point $x\in M$ the differential $df_x: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $df_x(T_xM)$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

One way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, let's take $U = \Bbb R^2$ and defined $r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)

At each point $x\in M$ the differential $df_x: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $df_x(T_xM)$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

One way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, if we can take $U = \Bbb R^2$ and defined $r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y)$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)

At each point $x\in M$ the differential $df_x: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $df_x(T_xM)$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

One way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, if we can take $U = \Bbb R^2$ and defined $r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)

At each point $x\in M$ the differential $df_x\: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $f_*T_xM$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

On way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, if we can take $U = \Bbb R^2$ and defined $r\: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y)$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)

At each point $x\in M$ the differential $df_x: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T_{f(x)}N$ a vector in the subspace $df_x(T_xM)$. To have the correct properties, this vector should coincide with the given vector in $T_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?

For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).

One way to do this, which is not canonical, is to choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).

Added Later:

I didn't read the question as carefully as I should have.

The submitter's choice of $r: U \to M$ amounts to the choice of a smooth retraction of a tubular neighborhod of $f(M)$ to $f(M)$.

The space of such choices is contractble, but I doubt that there is a preferred basepoint in this space of choices.

Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.

Example: in the $\Bbb R^2$ example, if we can take $U = \Bbb R^2$ and defined $r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y)$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r_*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii).

(By the way, the retraction induces a splitting $f^*TN \cong TM \oplus \nu$ of the kind mentioned above.)