Let $f\colon\mathcal{E}_1\to\mathcal{E}_2$ be an epimorphism of locally free sheaves on a Noetherian scheme $X$. Then also $\ker(f)$ is a locally free sheaf.
Proof. For all $x\in X$ one has a short exact sequence $0\to\ker(f)_x\to\mathcal{E}_{1,x}\to\mathcal{E}_{2,x}\to0$ of $\mathcal{O}_{X,x}$-modules, where $\mathcal{E}_{1,x}$ and $\mathcal{E}_{2,x}$ are free. In particular $\mathcal{E}_{2,x}$ is a projective $\mathcal{O}_{X,x}$-module, thus this exact sequence splits id est $\ker(f)_x$ is a direct summand of a free $\mathcal{O}_{X,x}$-module and thus it is projective. On the other hand, the projective modules over local rings are free and hence $\ker(f)_x$ is a free $\mathcal{O}_{X,x}$-module. By [H, Proposition II.5.7 and Exercise II.5.7.(b)] one has the claim. Q.E.D. $\Box$
Question: if $\mathcal{E}_1$ is coherent and $\mathcal{E}_2$ is locally free, then is $\ker(f)$ locally free?
My doubt: following the previous proof, under the new assumptions, I prove that $\mathcal{E}_1$ has to be locally free. Am I wrong?
[H] R. C. Hartshorne (1977) Algebraic Geometry, Springer-Verlag.
Your argument only shows that $\mathrm{ker}(f)_x$ is a direct summand of $\mathcal{E}_{1,x}$, so if $\mathcal{E}_1$ is not locally free, then $\mathrm{ker}(f)_x$ does not have to be projective.
And to get a simple example just assume that $\mathcal{E}_2 = 0$.