Turning a geometric vector bundle into a free $\mathscr{O}_X$-module

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Given a geometric vector bundle $V\to X$ on a scheme $X$, consider the sheaf of sections $\mathscr{I}(V/X)$ defined by $U\mapsto \operatorname{Hom}_U(U,V_{|U})$. The goal is to equip the sheaf of abelian groups $\mathscr{I}(V/X)$ an $\mathscr{O}_X$-module structure.

The idea is clear, take a trivializing affine cover $\{U_i = \operatorname{Spec} R_i\}$ of $X$. Over each $U_i$, we have $$ \mathscr{I}(V/X)(U) = \operatorname{Hom}_{U_i}(U_i,V_{|U_i}) \cong \operatorname{Hom}_{U_i}(U_i, \mathbb{A}_{U_i}^n) = \mathscr{O}_{X|U_i}^n $$ which has an obvious $\mathscr{O}_{X|U_i}$-module structure.

How do we glue the $\mathscr{O}_{X|U_i}$-module structures to an $\mathscr{O}_X$-module structure on $\mathscr{I}(V/X)$?

The problem is over the intersections $U_{ij} = U_i\cap U_j$, the two trivializing isomorphisms coming from $U_i$ and from $U_j$ give two distinct but isomorphic $\mathscr{O}_{X|U_{ij}}$-module structures on the same abelian group $\mathscr{I}(V/X)(U_{ij})$. But intuitively I can only fix one of them.

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Your vector bundle comes with a map $\mathbb{A}^1\times V\to V$ over $X$, that corresponds to scaling on fibers. You can use this scaling to define the action of $\mathcal{O}_X$ on sections of $V$.

Define a map of sheaves $\alpha\colon\mathcal{O}_X\times \mathcal{J}\to \mathcal{J}$ as follows: given sections $f\in \mathcal{O}_X(U)$ and $\phi\in \mathcal{J}(U)$, look at $f$ as a morphism $U\to \mathbb{A}^1$ and at $\phi$ as a morphism $U\to V|_U$. Set $\alpha(f,\phi)$ to be the section of $V|_U\to U$ determined by $U\stackrel{f\times\phi}{\longrightarrow} \mathbb{A}^1\times V|_U\to V_U$.

You can check that this defines a structure of $\mathcal{O}_X$-module on $\mathcal{J}$, and that this coincides with the ones you find using trivializations.