Given the following short exact sequence of holomorphic vector bundles on a compact Riemann surface:
$0\rightarrow M\rightarrow E \rightarrow N\rightarrow 0$
Fix a hermitian metric on $E$ and $n=rank(E)$. Let $\nabla$ be the corresponding unitary connection on the $U(n)-$bundle underlying $E$. Then it splits as
$\nabla=\begin{pmatrix}\nabla_M & C \\ B & \nabla_N\end{pmatrix}$
where $\nabla_M=pr_M\nabla$ where $pr_M$ denotes the orthogonal projection with respect to our Hermitian metric.
Since $M$ is a holomorphic subbundle of $E$ and $\nabla$ is compatible with $E$ (i.e. $\nabla^{0,1}=\overline{\partial}_E$), $\nabla_M$ is compatible with $M$. Hence $B$ is a ($1,0$)-form.
Why must we have $C=-B^\dagger$ (maybe with respect to some orthonormal basis)?
Is $\nabla$ being unitary equivalent to the matrix of $\nabla$ above being Hermitian with respect to an orthonormal basis? If this is true, maybe the explanation of $C=-B^\dagger$ has to do with it? Thank you.
$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Your suspicion (in a local unitary frame, a connection matrix is Hermitian) is correct. To expand on Mike Miller's comment, the point follows from a fundamental idiom of differential geometry: If $X$ and $Y$ are orthogonal sections of some Riemannian vector bundle (or, generally, if $\Brak{X, Y}$ is constant) and if $\nabla$ is compatible with the metric, the Leibniz rule and the fact that covariant differentiation commutes with contraction give $$ 0 = d\Brak{X, Y} = \Brak{\nabla X, Y} + \Brak{X, \nabla Y}. $$ In your Hermitian setting, this becomes $$ \Brak{BX, Y} = \Brak{\nabla X, Y} = -\Brak{X, \nabla Y} = -\Brak{X, CY} $$ for all local sections $X$ of $M$ and all local sections $Y$ of $M^{\perp}$.
Picking local unitary frames for $M$ and $M^{\perp}$, we have $B = -C^{\dagger}$ as (local) matrix-valued functions.