There are still strictly speaking elements floating around in the following since we are using indexing sets but maybe it is better? Consider for a set $S$ and an abelian group $A$ the isomorphisms $$Hom_{\mathrm{Set}}(S,UA) \cong Hom_{\mathrm{Set}}(\coprod_S \{*\},UA) \cong \prod_S UA$$ where $U$ is the forgetful functor. Since $U$ creates limits and we want $F$ to be left adjoint to $U$ the free abelian group $F(S)$ is the corepresenting object for the functor sending $A$ to $\prod_S A$. This is really just a reinterpretation of the fact that we want to hit things with the free abelian group functor but maybe it is closer to what you are looking for?
Also I am not completely sure what you mean by your last comment - if you are trying to understand the relationship between homology and cohomology via the chain/cochain complexes you might be after the following?
Theorem Let $R$ be a ring and let $P$ be a chain complex of projective $R$-modules such that each $d(P_i)$ is also projective. Then for every $i$ and every $R$-module $N$, there is a non-canonically split short exact sequence $$ 0 \to Ext^1_R(H_{i-1}(P),N) \to H^i(Hom_R(P,N)) \to Hom_R(H_i(P),N) \to 0$$
So in particular for complexes of free abelian groups one can apply this and it tells you how taking duals alters the cohomology.
I'd like to second Ryan Budney's answer about the difference being the maps in the finite rank setting. Although unless you have a natural setting in which you are considering chains and cochains (for instance singular homology/cohomology) I still don't quite understand exactly what you are after since you can always reindex to move between them.