Let $(X,\mathcal O_X)$ be the affine scheme of a commutative Noetherian ring $R$. Let $J$ be an ideal of $R$. From the sheaf $\mathcal O_X$ and the closed subset $Z:=V(J)$ of $X$ , how do we recover the sheaf of $\mathcal O_X$-modules on $X$ associated to the $R$-module $R/J$ ? Is it $\mathcal O_X|_Z:=i^{-1}(\mathcal O_X)$ or $i^*(\mathcal O_X)$ (where $i:Z \to X$ is the inclusion) or something else ?
(In case it is unclear, by the sheaf associated to an $R$-module $M$, I mean the sheaf which takes every basic open set of the form $D(f)$ of $X$ to the localization $M_f$ )
NOTE: I guess strictly speaking, neither $\mathcal O_X|_Z$ nor $i^*(\mathcal O_X)$ is a Sheaf on $X$ but rather on $Z$. By the comment of user Ruben du Burck below it seems like the sheaf on $X$ I'm looking for is $i^*(\mathcal O_X)$ extended to $X$ by zero outside of $Z$ i.e. just $i_*(i^*(\mathcal O_X))$. But I'm still on the lookout for a definite answer. Thanks.
You cannot recover $R/J$ from just the sheaf $\mathcal{O}_X$ and the set $Z$; if you could, that would mean any two ideals $J,J'$ with $V(J)=V(J')$ are the same, but this is not true.
If you know $Z$ not just as a closed subset of $X$ but as the closed subscheme assoociated to the ideal $J$, then the sheaf on $X$ you seek is just $i_*(\mathcal{O}_Z)\cong i_*(i^*(\mathcal{O}_X))$. This follows immediately from the fact that if $\varphi:A\to B$ is a homomorphism of rings with corresponding morphism of schemes $f:\operatorname{Spec} B\to \operatorname{Spec} A$, then the operation $f_*$ on quasicoherent sheaves corresponds to taking a $B$-module and considering it as an $A$-module via $\varphi$, and the operation $f^*$ corresponds to $B\otimes_A -$. In this case, $A=R$ and $B=R/J$, so to get $R/J$ as an $R$-module, you just take $i_*(\mathcal{O}_Z)$, and $\mathcal{O}_Z$ is also the same as $i^*(\mathcal{O}_X)$ since $R/J\otimes_R R\cong R/J$.