Let $f\colon (X,\mathcal{O}_X)\rightarrow (Y,\mathcal{O}_Y)$ be a map of ringed spaces. In the Stacks Project, in the proof of Tag 09U8, they basically claim at some point, that for any flat $\mathcal{O}_Y$-module $\mathcal{F}$ its pullback $f^*\mathcal{F}$ is also flat. Now they do link to a proof of this result, but in that linked statement they assume $f$ to be a flat morphism.
So if $f$ is flat, that the statement is clear, but does it also hold for an arbitrary $f$?
The trick is that you can change what $\mathcal{O}_X$ is to get a flat morphism. Let $X'=(X,f^{-1}\mathcal{O}_Y)$, which is flat over $Y$, as $(f^{-1}\mathcal{O}_Y)_x\cong\mathcal{O}_{Y,f(x)}$ and the induced map $\mathcal{O}_{Y,f(x)}\to\mathcal{O}_{Y,f(x)}$ is an isomorphism. Writing $$f^*\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{G} \cong f^{-1}\mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X \otimes_{\mathcal{O}_X} \mathcal{G} \cong f^{-1}\mathcal{F}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{G}$$ for $\mathcal{G}$ a sheaf of $\mathcal{O}_X$-modules, we see that pulling back to $X$ is equivalent to pulling back to $X'$ on the level of sheaves of abelian groups, and $X'\to Y$ is flat, so we can use the result.
This is essentially covered at the last statement (0GMU) in 02N2, but it could be added to the lemma you're reading to make it a little clearer, perhaps.