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).