Assume $S$ is a smooth projective surface and $E$ is a nontrivial torsion free sheaf of rank $r$ on $S$ that comes with a surjective morphism $\mathcal{O}_S^{\oplus t}\rightarrow E\rightarrow 0$ for some $t>r$.
$\textbf{Question:}$ Is it always possible (depending on $S$) to find a free subsheaf $\mathcal{O}_S^{\oplus r+1}\hookrightarrow \mathcal{O}_S^{\oplus t}$ such that the composition $\mathcal{O}_S^{\oplus r+1}\rightarrow E$ is surjective, except at maybe finitely many points in $S$?
This result is stated for a $K3$-surface (without proof) in Mukai's "Duality of polarized K3 surfaces" on page 320. So maybe this is something very obvious? But I was unable to prove this for myself. If it is obvious I am happy just to see a refrence in the literature. Otherwise how to prove this?
Yes, this is true; in fact, this is a version of the Bertini theorem.
To prove consider the Grassmannian $\mathrm{Gr}(r+1,t)$ with the tautological subbundle $\mathcal{U} \subset \mathcal{O}^{\oplus t}$. Further, consider the product $S \times \mathrm{Gr}(r+1,t)$ and the composition $$ \mathcal{O} \boxtimes \mathcal{U} \hookrightarrow (\mathcal{O} \boxtimes \mathcal{O})^{\oplus t} \twoheadrightarrow E \boxtimes \mathcal{O}. $$ Let $D \subset S \times \mathrm{Gr}(r+1,t)$ be its degeneration locus (i.e., the locus where the composition is not surjective). The fiber $D_s$ of the projection $D \to S$ over a point $s \in S$ parameterizes all $(r+1)$-dimensional subspaces in the $t$-dimensional space $\Bbbk^t$ that intersect the $(t - r)$-dimensional kernel $\mathrm{Ker}(\Bbbk^t \twoheadrightarrow E_s)$ in a subspace of dimension $\ge 2$. Therefore, $$ \dim(D_s) = 2(t - r - 2) + (r-1)(t-r-1) = (r+1)(t-r-1) - 2. $$ Thus, $D_s$ has codimension 2 in the fiber of $S \times \mathrm{Gr}(r+1,t)$ over $s \in S$, hence $D$ has codimension 2 in the product. Now it follows that for a general point $U \subset \mathrm{Gr}(r,t+1)$ the fiber $D_U$ of $D$ over $U$ has codimension 2. It remains to note that $D_U$ is precisely the locus of points of $S$, where the composition $$ U \otimes \mathcal{O} \hookrightarrow \mathcal{O}^{\oplus t} \twoheadrightarrow E $$ is not surjective.