Consider an extension $0\rightarrow L \overset{\alpha}{\rightarrow} E \overset{\beta}{\rightarrow} L' \rightarrow 0$ of bundles and bundle homomorphisms, where $L$ and $L'$ are line bundles. (Let's assume that all bundles and maps are holomorphic.)
Does it split locally? i.e. is there an open cover $\{U_i\}$ and maps $\sigma_i: L'\vert_{U_i} \rightarrow E\vert_{U_i}$ such that $\beta\circ \sigma_i = \text{id}$?
You could use a common trivialization for $L$, $L'$, and $E$, and then you get an extension of trivial bundles, say $0\rightarrow U\times \mathbb{C} \rightarrow U\times \mathbb{C}^2 \rightarrow U\times \mathbb{C} \rightarrow 0$. So a related question is: do extensions of trivial bundles split locally (or even globally)?
In the smooth case you can use a partition of unity to construct a Hermitian metric, and then you get splitting globally. I was hoping to get this local splitting for vector bundles in any setting, i.e. smooth, holomorphic, algebraic, etc.
I'll just shortly elaborate about the algebraic proof. In fact line bundles are in one-to-one correspondence with locally free sheaves of rank 1. They are locally given by projective modules (locally free = projective), and then the proof is just a simple algebraic fact that a surjection onto a projective modules splits, which is just the definition of projective.