This condition seems rather strong so I am not sure if this is true. We assume $X$ is (locally) noetherian and $\mathscr{F}, \mathscr{G}$ are locally free of finite rank. Now, I claim that $\mathscr{H} = \ker(\mathscr{F} \to \mathscr{G})$ is locally free. Hartshorne's exercise II.5.7 states that a coherent sheaf $\mathscr{H}$ on $X$ is locally free iff $\mathscr{H}_x$ is free over $\mathcal{O}_x$ for all $x \in X$.
Now, let $x \in X$ be arbitrary. Since $\mathscr{H}_x$ fits into the exact sequence below $$0 \to \mathscr{H}_x \to \mathcal{O}_x^{\oplus n} \to \mathcal{O}_x^{\oplus m} \to 0.$$ Now, since $\mathcal{O}_x^{\oplus n}, \mathcal{O}_x^{\oplus m}$ are projective $\mathcal{O}_x$-modules, this exact sequence splits so that $\mathscr{H}_x \oplus \mathcal{O}_x^{\oplus m} \cong \mathcal{O}_x^{\oplus n}$ so that $\mathscr{H}_x$ is a direct summand of a free module and hence projective. Now, $\mathscr{H}_x$ is projective over a local ring $\mathcal{O}_x$ and is hence free.
The above seems fine. I am mostly concerned with the case that $X = \operatorname{Spec} A$, where this would imply that the kernel of a surjective map of free modules of finite rank is free. Based on this answer, this does not appear to be true. Am I making a mistake above or am I mistaken in assuming a locally free sheaf on an affine scheme is free?
Best Answer
Your argument before "seems fine" is actually fine. Your mistake is assuming that a locally free sheaf on an affine scheme must be free - the real result is that a locally free sheaf on an affine scheme corresponds to a projective module, and not all projective modules are free. For instance, $(2,1+\sqrt{-5})$ is a projective module of rank 1 over $\Bbb Z[\sqrt{-5}]$ which is not free.