Let $\pi:Z\to B$ be an elliptic fibration over some base $B$. Suppose two vector bundles $V$ and $V'$ differ only in that on each fibre (i.e. elliptic curve) they are related by some deformation, i.e. they sit in the same extension group.
My question is whether it follows that $V$ and $V'$ are themselves related by a deformation. I feel that this should obviously be the case, but I don't have a solid enough understanding of extensions to convince myself.
If the base $B$ is local, or affine with trivial Picard group, then maybe you can extend the arguments I gave in my answer to your other question here. But as soon as $B$ is global, say a projective curve, then you are dealing with bundles on a surface. They have a first Chern class that is a divisor on the surface, which may not be uniquely determined by its restrictions to a pencil of curves. There is also a second Chern class, which cannot be seen on a curve. For a specific example, take a bundle on your elliptic curve $E$, and consider its pullback $V$ to $E \times \mathbb{P}^1$ (the projective line). Then take $V'$ to be $V$ tensor the pullback of $\mathcal{O}(1)$ on $\mathbb{P}^1$. In this case, $V$ and $V'$ have isomorphic restrictions to every elliptic fiber, but cannot be deformation equivalent since their first Chern classes are different.