In Vakil's notes page 76, he claims that a sheaf of abelian groups is the same as a $\underline{\mathbb{Z}}$-module, where $\underline{\mathbb{Z}}$ is the constant sheaf associated to $\mathbb{Z}$.
Let $F$ be a sheaf of abelian group over $X$. Given an open set $U\subset X$, we need to find an action
$$\underline{\mathbb{Z}}(U)\times F(U)\to F(U),$$ that is compatible with restriction. Recall that $$\underline{\mathbb{Z}}(U) = \{f: U\to \mathbb{Z}: \mathbb{Z} \text{ has discrete topology, $f$ continuous.}\}$$ So how does one define $f\times \alpha\mapsto ?$
I highly doubt the statement is correct. Idealy since an abelian group is a $\mathbb{Z}$-module, don't we want the constant presheaf asscoiated to $\mathbb{Z}$ rather than the constant sheaf?
I worked things out and have the following conclusion.
Abelian groups are the same as $\mathbb{Z}$-modules, therefore sheafs of abelian groups are the same as $\underline{\mathbb{Z}}^{pre}$-modules. However, due to the fact that sheafs have the identity and gluability property, this $\underline{\mathbb{Z}}^{pre}$ action can be canonically extended to a $\underline{\mathbb{Z}}$ action, the way @hoot described, where $\underline{\mathbb{Z}}^{pre}$ is identified as the constant functions inside $\underline{\mathbb{Z}}$.
This can also be explained from the adjunction of embedding and sheafification as @Zhen Lin described. The fact that the sheafification of the constant presheaf is the constant sheaf implies that a $\underline{\mathbb{Z}}^{pre}$-module extends uniquely to a $\underline{\mathbb{Z}}$-module.