When does a morphism $f:E'\to E$ of vector bundles over the same base $B$ make $E'$ a vector bundle over $E$? Definitely $f$ has to be surjective, and this seems like it is also sufficient. Because any point of $B$ has a neighborhood $U$ over which $E'$ and $E$ are both trivial. Then it seems to me that $E'$ is trivialized over the open set $p^{-1}(U)\subseteq E$, where $p$ is the projection $E\to B$.
(Maybe it's a simple question, but just wanted to double check).
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'\cong E\oplus \ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $\ker(f)\to B$ along $E\to B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.