Let $X$ be a scheme, $\mathcal{E}$ a locally free $\mathcal{O}_X$-module of finite rank and $p: E\to X$ the corresponding geometric vector bundle (with global sections $\mathcal{E}$). Do we have an exact sequence of $\mathcal{O}_E$-modules like this:
$$0\to p^*\mathcal{E}^\vee\to\mathcal{O}_E\to s_*\mathcal{O}_X\to 0?$$ Or do these 3 terms somewhat related?
Here $s: X\to E$ is the zero section, $\mathcal{E}^\vee$ the dual. Anyway, I mainly want to know the case when $\mathcal{E}$ is an invertible sheaf.
Update: Calculating locally, we find that it's indeed the case for line bundles. In higher rank case, we still have an exact sequence $p^*\mathcal{E}^\vee\to\mathcal{O}_E\to s_*\mathcal{O}_X\to 0$. (One must be very careful in writing down the relevant maps even in the local case, though.)
I think the following is the best you can do.
If $P\to X$ is a $G$ bundle then its pullback along itself $P\times_XP\to P$ is trivial. Proof: the diagonal map $P\to P\times_XP$ gives a section.
This is not true for vector bundles $E\to X$ (as "trivial iff there is a section" is no longer true). However, what you can say is that you have a tautological section of $p^*E=E\times_XE\to E$ given by the diagonal map $E\to E\times_XE$. This section vanishes on $X\subseteq E$. Call the section $s$.
We also have the evaluation map $\mathscr{E}^\vee\otimes\mathscr{E}\to\mathscr{O}_X$, and pulling back gives $p^*\mathscr{E}^\vee\otimes p^*\mathscr{E}\to \mathscr{O}_E$.
Putting this together, you get a map of vector bundles on $E$ $$p^*\mathscr{E}^\vee \ \stackrel{\text{id}\otimes s}{\longrightarrow}\ p^*\mathscr{E}^\vee\otimes p^*\mathscr{E}\ \longrightarrow \ \mathscr{O}_E.$$ However, since $s$ isn't non-vanishing this is not surjective, even in the line bundle case. And it's certainly not injective when $\text{rk}\mathscr{E}>1$.