We know that a morphism $\phi$ between two presheaves $\mathcal {F}$ and $\mathcal {G}$ on a topological space $X$ is the data of maps $$\phi(U):\mathcal {F}(U) \to \mathcal {G}(U)$$ $\forall \ U$ open in $X$ such that the following diagram
$\require{AMScd}$ \begin{CD} \mathcal {F}(V) @>\phi(V)>> \mathcal {G}(V)\\ @V Res_{V,U} V V @VV res_{V,U} V\\ \mathcal {F}(U) @>\phi(U)>> \mathcal {G}(U) \end{CD} commutes for every inclusion $U \subset V$ in $X$. Now , if $\mathcal {F}$ and $\mathcal {G}$ are $\mathcal O_X$ modules (where $(X, \mathcal O_X)$ is a ringed space) then morphisms between them are essentially the data of abelian group morphisms $$\phi(U):\mathcal {F}(U) \to \mathcal {G}(U)$$ $\forall \ U$ open in $X$ such that for them ,the above diagrams commute for each inclusion. Now everywhere it is written that this morphism is same as the data of maps$$\phi(U):\mathcal {F}(U) \to \mathcal {G}(U)$$ $\forall \ U$ open in $X$ which are $\mathcal O_X(U)$ module morphisms. But I have tried a number of times to prove that the property of being abelian group homomorphism together with the compatibility of restriction maps (which we get by virtue of $\mathcal F$ and $\mathcal G$ being $\mathcal O_X$ modules) implies that $\phi(U)$'s are $\mathcal O_X$ module morphisms and conversely if we have a bunch of $\mathcal O_X$ module morphisms from $\mathcal F(U)$ to $\mathcal G(U)$ ,then automatically the aforesaid diagram will commute giving rise to morphism from $\mathcal F$ to $\mathcal G$, and I have failed to see the reason why it should be so.The main difficulty is taking $a\in \mathcal O_X(U)$ out of the expression $\phi(U)(a.f)$ for any $f\in \mathcal F(U)$(I have mainly used compatibility condition to show that those abelian group morphisms are indeed appropriate module morphism). I hope once this part is clear, the converse will be easier to see.
Any help from anyone is welcome.