Let $S^k$ be the presheaf on a space $X$ that assigns to every open set $U$ the abelian group $S^k(U)$ of singular k- cochains on $U$. This is clearly not a sheaf. Consider the sheafification $F^k$ of each $S^k$. These sheaves form an exact resolution of the constant sheaf of integers.
We can take global sections on this sheaf resolution to obtain a cochain complex $F^*(X)$. Does the cohomology of this cochain complex coincides with ordinary singular cohomology?
You haven't said this explicitly, but your claim is that the sheaf cohomology of the constant sheaf $\Bbb Z_X$ is isomorphic to the singular cohomology of $X$: $$H^\ast(X, \Bbb Z_X) \cong H^\ast_{\text{sing}}(X; \Bbb Z).$$ This is true for $X$ locally contractible. One checks that the complexes $S^\ast(X)$ and $\Gamma(X, S^\ast(X))$ are quasi-isomorphic. More generally, the assertion holds for any abelian group $G$: $$H^\ast(X, G_X) \cong H^\ast_{\text{sing}}(X; G)$$ when $X$ is locally contractible.
I don't have any references on hand, but I would expect that you could find the proof in most books that introduce the concept of sheaf cohomology.