There are (at least) two details you are missing.
(1) This is not an equivalence of categories between finitely generated graded modules and coherent sheaves. If your module is $0$ in all sufficiently large degrees, then the corresponding sheaf will be zero. For example, let $S=k[x,y]$ and $M=S/\langle x,y \rangle$. The sheaf on $\mathbb{P}^1$ corresponding to $M$ is the zero sheaf.
The category you want to work with is the one whose objects are finitely generated graded modules, and where we formally invert any map which is an isomorphism in sufficiently large degree. (Alternative formulation: we formally invert a map $f:M \to N$ if, for any $s$ in the irrelevant ideal, $s^{-1} f: s^{-1} M \to s^{-1} N$ is an isomorphism.)
(2) Not every coherent sheaf is a vector bundle. Correspondingly, not every finitely generated graded module will correspond to a vector bundle. If we were dealing with an affine variety, vector bundles would correspond to locally free modules. (Also called projective modules.) For projective varieties, things are a little trickier: the criterion is to be locally free away from the irrelevant ideal.
Sadly, I believe that there exist examples of modules which are locally free away from the irrelevant ideal, but are not isomorphic (in the above category) to any module which is locally free at the irrelevant ideal. This should be related to my question here. But you can go a long while without paying attention to this detail.
The answer is positive. Let $P$ be the principal $\mathrm{GL}_n$-bundle associated with $E$; then the space of flags is the quotient $P/B$, where $B$ is the Borel subgroup of $\mathrm{GL}_n$ consisting of upper triangular matrices. Set $Z = P/T$, where $T$ is the maximal torus consisting of diagonal matrices. A point of $Z$ is a point of $X$, plus $n$ independent 1-dimensional linear subspaces of the fiber of $E$. The projection $Z \to P/B$ is a fibration with contractible fibers, hence the pullback from the cohomology of $P/B$ to that of $Z$ is an isomorphism. Since the cohomology of $X$ injects into the cohomology of $P/B$, it also injects into the cohomology of $Z$.
Best Answer
There are two approaches to this question, one by Toledo and Tong (using Cech covers or hypercovers) and one by Block (using the Dolbeaut algebra). Block's paper is at http://arxiv.org/abs/math/0509284