Pullback of modules is right exact, and flat stalk maps make it exact

Let ${\mathcal{M}}^{\mathrm{\prime }}\to \mathcal{M}\to {\mathcal{M}}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ be an exact sequence of ${\mathcal{O}}_Y$-modules, and fix $x\in X$ with $y=f(x)$. By [L1], the stalk sequence after applying $f^{-1}$ is ${\mathcal{M}}_y^{\mathrm{\prime }}⟶{\mathcal{M}}_y⟶{\mathcal{M}}_y^{\mathrm{\prime }\mathrm{\prime }}⟶0,$ which is exact because the original sequence is exact at every stalk. Hence $f^{-1}$ is exact on the underlying sheaves.

By [F1] and [L2], the stalk of the pulled-back sequence at $x$ is ${\mathcal{O}}_{X,x}{\otimes }_{{\mathcal{O}}_{Y,y}}{\mathcal{M}}_y^{\mathrm{\prime }}⟶{\mathcal{O}}_{X,x}{\otimes }_{{\mathcal{O}}_{Y,y}}{\mathcal{M}}_y⟶{\mathcal{O}}_{X,x}{\otimes }_{{\mathcal{O}}_{Y,y}}{\mathcal{M}}_y^{\mathrm{\prime }\mathrm{\prime }}⟶0.$ Step 1.1 gives exactness before tensoring, so [L4] makes this sequence right exact. Since this holds for every $x$, [L3] shows that $f^{\ast }$ is right exact.

Assume every stalk ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ is flat. By [F2], for each $x\in X$ the functor ${\mathcal{O}}_{X,x}{\otimes }_{{\mathcal{O}}_{Y,f(x)}}-$ is exact on ${\mathcal{O}}_{Y,f(x)}$-modules. Applying this to the exact stalk sequence from step 1.1 shows that the sequence in step 2.1 is exact at every $x$. Therefore [L3] implies that $f^{\ast }$ is exact.