Only a partial answer, starting with the disclaimer that it does not apply when the base is the total space of a vector bundle on a projective curve.
''If $X$ is a Stein manifold of dimension $2$, then each holomorphic vector bundle $V \to X$ of rank $r >1$ has a trivial one-dimensional subbundle''. In particular, it fits into a short exact sequence.

Proof: $X$ contains a $2$-dimensional CW-complex $K \subset X$ as a deformation retract. A generic (smooth) section of $V$ has a zero set of real dimension $4-2r < 2$, and moreover the zero set does not meet $K$ for dimensional reasons and by transversality. So $V|_K$ has a trivial complex line bundle as subbundle.

By homotopy invariance of vector bundles, this shows that $V$ has a trivial smooth subbundle. Now study the holomorphic fibre bundle $Mon(\mathbb{C};V)\to X$ (a point over $x $ is a complex linear monomorphism $\mathbb{C} \to V_x$). The fibre is the complex homogeneous space $Mon(\mathbb{C};\mathbb{C}^r)$, which is the quotient of $GL_r (\mathbb{C})$ by the stabilizer subgroup $G$ of the action of $GL_r (\mathbb{C})$ on $\mathbb{C}^{r} \setminus 0$.

The first paragraph says that there is a global smooth section $X \to Mon(\mathbb{C};V)$. Finding a holomorphic subbundle is the same as finding a holomorphic section.

Grauerts theorem (''Analytische Faserungen \"uber holomorph-vollst\"andigen R\"aumen'') says that for bundles of the above type over Stein manifolds, each smooth section is homotopic to a holomorphic section.

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