In Görtz-Wedhorn, the argument used to prove that the pullback of a quasi-coherent sheaf $\mathcal{G}$ by the map $f: (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ is a quasicoherent sheaf is in remark 7.23, and goes as follows:
"Let $f: (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ be a morphism of ringed spaces. If $\mathcal{G}$ is a quasi-coherent module on $\mathcal{O}_Y$ it follows from $f^*(\mathcal{O}_Y) = \mathcal{O}_X$ and from the fact that $f^*$ is right exact and commutes with direct sums that $f^*(\mathcal{G})$ is a quasi-coherent $\mathcal{O}_X$ module."
I thought that his argument is meant to go as follows: because $\mathcal{G}$ is quasi-coherent, then for each $y \in Y$, there exists an open neighborhood $V \subseteq Y$ and sets I, J such that $$\mathcal{O}_Y|_V ^{(J)} \to \mathcal{O}_Y|_V^{(I)} \to \mathcal{G} \to 0$$ Then, for each $x \in X$, consider a neighborhood $V$ of $f(x)$ with this property. We have $f|_{f^{-1}(V)}^*(\mathcal{O}_Y|_V^{(J)}) = \mathcal{O}_X|_{f^{-1}(V)}^{(J)}$ - for the set $I$, analogously. Applying $f^*$ to the above exact sequence would give us: $$\mathcal{O}_X|_{f^{-1}(V)} ^{(J)} \to \mathcal{O}_X|_{f^{-1}(V)}^{(I)} \to f^*\mathcal{G} \to 0$$ And this is an exact sequence, because $f^*$ is right exact.
Is this what the author meant? Because it seems to me that there is something wrong in this reasoning: the fact that $f^*$ is a contravariant functor would mean that the sequence would have its arrows reversed. If the author meat otherwise, what did he mean? Can anyone shed some light?