Just wish to clarify, is it true that in order to show some vector bundles (over the same space) fit into a short exact sequence we just need to check that their fibers fit into a short exact sequence of vector spaces?
Thank you very much for your attention!
No, this is not true. Obviously it is necessary for the fibers to all fit into short exact sequences, but it is not sufficient.
For a counterexample, let $X$ be any space that has a nontrivial vector bundle $E$ of rank $n$. Let $F$ denote the trivial vector bundle over $X$ of rank $n$, and let $0$ denote the vector bundle of rank $0$ over $X$. Then for each $x\in X$, there is a short exact sequence $0\to 0_x\to E_x\to F_x\to 0$ of fibers at $x$, since $E_x$ and $F_x$ are both $n$-dimensional vector spaces and $0_x$ is the trivial vector space. But there is no short exact sequence of vector bundles $0\to 0\to E\to F\to 0$, since this would imply $E\cong F$, but $E$ is nontrivial.
However, as Mariano Suárez-Alvarez commented, if you already have maps of vector bundles $E\to F$ and $F\to G$, then it is true that the sequence $0\to E\to F\to G\to 0$ is exact iff $0\to E_x\to F_x\to G_x\to 0$ is exact for each $x\in X$. I can't prove this without knowing what your definition of "exact sequence of vector bundles" is (indeed, one such definition is that a sequence is exact iff it is exact on each fiber).
