As pointed out in the comments, every characteristic class in $H^d(BO(d), G)$ provides a $G$-valued locally computable invariant of $d$-manifolds, by pulling back via the classifying map of the tangent bundle. As argued by Levitt and Rourke, there are local combinatorial formulas on trianglulations for dual simplicial $d-i$-cycles representing any degree-$i$ characteristic classes, whose values on a $d-i$-simplex only depend on the star of that simplex. To be precise, for this to work we need to add some decorations to the triangulation, namely a local ordering.
So every degree-$d$ characteristic class yields a local combinatorial formula for a simplicial $0$-cycle, whose summation over vertices is an invariant of the type in the question. The "objects" are the different stars of vertices, to which we associate elements of $G$ according to the local formula. Summing up all the $G$-values on the vertices corresponds to evaluating the invariant.
Concrete local formulas are in fact known for many cases. For $G=\mathbb{Z}_2$, the relevant characteristic classes are generated by degree-$d$ polynomials (with $\mathbb{Z}_2$ sum and cup product) of Stiefel-Whitney classes. For the latter, combinatorial formulas are given by Goldstein and Turner, and a combinatorial cup product for cycles can be easily obtained via its geometric interpretation as intersection. For $G=\mathbb{Z}$ on oriented manifolds, we additionally need the Bockstein homomorphism for the short exact sequence
$$\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}_2$$
and the Pontryagin classes. The former only involves simplex-wise operations and the co-boundary operator. The Pontryagin classes are the only case for which no satisfactory local formulas exist to date. $\mathbb{Q}$-valued formulas have been described by Gaifullin, but not very explicitly.
I'm not aware of any converse argument that every local formula for a $0$-cycle invariant corresponds to a characteristic class, however I don't think there are any known examples for invariants which don't.