Let $\left(T,\mathcal{O}_{T}\right)$ and $\left(S,\mathcal{O}_{S}\right)$ be schemes and $f:T\rightarrow S$. $U\subset S$ and $V\subset T$ are some open sets that correspond to the definition of $f^{*}:\mathcal{O}_{S}^{n}\rightarrow \mathcal{O}_{T}^{n}$.
I refer to the following definition of $f^{*}$ (with $X$ instead of $S$ and $Y$ instead of $T$):
For all pairs $\left(U,V\right)$ formed from open sets $U$ of $X$ and $V$ of $Y$ such that $f\left(V\right)\subset U$, exists a morphism of rings $f^{*}:\mathcal{O}_{X}\left(U\right)\rightarrow\mathcal{O}_{Y}\left(V\right)$, with the requirement that if $U$ and $U^{\prime}$ are two open sets of $X$, and $V$ and $V^{\prime}$ are two open sets of $Y$ with $V^{\prime}\subset V$, $f\left(V\right)\subset U$ and $f\left(V^{\prime}\right)\subset U^{\prime}$, then the diagram
commutes.
How can it be proved that $$ q\left(U\right):\mathcal{O}_{S}^{n}\left(U\right)\rightarrow\mathcal{O}_{S}^{n}\left(U\right)\textrm{ surjective}\Rightarrow \left(f^{*}q\right)\left(U\right):\mathcal{O}_{T}^{n}\left(V\right)\rightarrow\mathcal{O}_{T}^{n}\left(V\right)\textrm{ surjective}? $$ Here $q$ is an endomorphism between $\mathcal{O}_{S}^{n}$ and $\mathcal{O}_{S}$-module $\mathcal{O}_{S}^{n}$.
I already proved the converse.
Am I missing something here?
Edit: it looks like I should use the fact that $f^{*}\left(\mathcal{O}_{S}^{n}\right)=\mathcal{O}_{T}^{n}$ (vs. merely $f^{*}\left(\mathcal{O}_{S}^{n}\right)\subseteq \mathcal{O}_{T}^{n}$).
The question is unclear as stated (what is $q$? what's the purpose of $s,t$? what does "correspond to the definition of $f^{*}:\mathcal{O}_{S}^{n}\rightarrow \mathcal{O}_{T}^{n}$" mean?). So here is what perhaps was meant:
Set $f : T \to S$ be a morphism of arbitrary ringed spaces. Let $q : \mathcal{O}_S^n \to \mathcal{O}_S^n$ be a homomorphism of $\mathcal{O}_S$-modules such that $q(U) : \mathcal{O}_S(U)^n \to \mathcal{O}_S(U)^n$ is surjective for some open subset $U$. Let $V \subseteq f^{-1}(U)$ be some open subset. Why is then $(f^* q)(V) : \mathcal{O}_S(V)^n \to \mathcal{O}_S(V)^n$ surjective?
First, I claim that actually $q|_U : \mathcal{O}_U^n \to \mathcal{O}_U^n$ is an isomorphism. In fact, $q(U)$ is an isomorphism since it is a surjective endomorphism of a f.g. free module. The determinant of $q(U)$ is therefore a unit global section of $\mathcal{O}_U$. This stays a unit global section if we restrict to open subsets $V$ of $U$, which shows that $q(V)$ is also an isomorphism.
Let $f_U$ denote the corestriction $f^{-1}(U) \to U$. Since $(f_U)^*$ is a functor, it maps isomorphisms to isomorphisms. Thus, $(f_U)^* q|_U : \mathcal{O}_{f^{-1}(U)}^n \to \mathcal{O}_{f^{-1}(U)}^n$ is an isomorphism. Now we may restrict this to an open subset $V \subseteq f^{-1}(U)$ and take sections on $V$ to see that $(f^* q)(V)$ is an isomorphism.
(Notice that we really need isomorphisms for this kind of reasoning; epimorphisms don't restrict to epimorphisms on sections, since sheaf cohomology is an obstruction for this.)