I read the following in "Sheaves on Manifolds by Kashiwara" on page 87
The relevant definition is
Let $\mathcal{R}$ be a sheaf of rings on $X$. An $\mathcal{R}$-module $M$ (or a sheaf of modules over $\mathcal{R}$ is a sheaf $M$ such that for each open set $U\subset X$, $M(U)$ is a left $\mathcal{R}(U)$ module, and for any $V\subset U$, the restriction morphism is compatible with the structure of module, that is $\rho_{V,U}(sm)=\rho_{V,U}(s)\cdot \rho_{V,U}(m)$ for any $s\in \mathcal{R}(U), m\in M(U)$. One denotes by $Mod(\mathcal{R})$ the category of left $\mathcal{R}$ modules.
The example they gave was the following. Let $Sh(X)$ be the category of sheaves with values in the category of abelian groups. Let $\mathbb{Z}_{X}$ be the sheaf associated to the presheaf $U\mapsto \mathbb{Z}$. Then we apparently have
$$Sh(X)=Mod(\mathbb{Z}_{X})$$
I don't understand how they got the last equality. $\mathbb{Z}_{X}$, as a sheaf, can be described as follows. for $U\subset X$ we have that $\mathbb{Z}_{X}(U)$ is the ring of locally constant functions from $U$ to $\mathbb{Z}$. Consider $\mathcal{F}\in Sh(X)$, then $\mathcal{F}(U)$ should be a $\mathbb{Z}_{X}(U)$ module. Let $f\in \mathbb{Z}_{X}(U)$ and let $s\in \mathcal{F}(U)$, then how does $f$ act on $s$? I can't see any obvious way to have $f$ act on $s$
The key facts here are:
Let $\mathbb Z_X^-$ be the presheaf $U \mapsto \mathbb Z$ and let $\mathbb Z_X$ be its sheafification. Let $\mathcal F$ be a sheaf of abelian groups over $X$. By (1), $\mathcal F$ is a $\mathbb Z_X^-$-module. The corresponding morphism $\mathbb Z_X^- \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F)$ gives rise to a compatible morphism $\mathbb Z_X \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F)$ by the universal property of sheafification. By (2), this morphism gives $\mathcal F$ its $\mathbb Z_X$-module structure.