I am following Kobayashi's book Differential geometry of complex vector bundles, which is proving a challenge to follow in detail, as many of the proofs leave up to the reader to fill in the gaps.
Preliminaries
My question is related to Section I.6 Subbundles and quotient bundles: here we have a rank $r$ holomorphic vector bundle $E\rightarrow M$ endowed with a Hermitian structure $h$ and a rank $p$ holomorphic subbundle $S \subset E$. We can define the quotient bundle $Q = E/S$ fitting in the short exact sequence $$ 0 \rightarrow S \rightarrow E \rightarrow Q \rightarrow 0$$ and $Q$ being $C^\infty$ isomorphic to the orthogonal complemente $S^\perp$. Consider the canonical Hermitian connection D (metric compatible and $D^{0,1} = \bar{\partial}$). We can decompose, taking sections with values in the subbundle
$$D \xi = D_S \xi + A \xi~~~~~~~~\forall \xi\in\Omega^0(S)$$ $$D \eta = D_{S^\perp} \eta + B \xi~~~~~~~~\forall \eta\in\Omega^0(S^\perp)$$
and it is easy to see that $D_S$ is the Hermitian connection in $S$, $A$ is a $Hom(S,S^\perp)$-valued (1,0)-form, $B$ is a $Hom(S^\perp,S)$-valued (0,1)-form, and furthermore, for $\xi \in\Omega^0(S),\eta\in\Omega^0(S^\perp)$ we have $$ h(A\xi,\eta) = - h(\xi,B\eta)$$
My problem
I want to deduce the analogues of Gauss-Codazzi equations for Hermitian vector bundles, i.e. that the curvature act like $$ R = \left(\begin{array}{cc} R_S - B\wedge B^* & D^{(1,0)}B\\ -D^{(0,1)}B^* & R_{S^\perp} - B^*\wedge B \end{array}\right) $$
Using frames for the $C^\infty$ decomposition $E = S\oplus S^\perp$ greatly helps. If we use the notational convention $1\leq a,b,c \leq p$, $p+1\leq \lambda,\mu,\nu \leq r$ it is very simple to split the indices (summing over repeated indices everywhere) and see that
$$DDe_a = (d\omega^b_a + \omega^b_c \wedge \omega^c_a + \omega ^b_\lambda \wedge \omega^\lambda_a)e_b + (d\omega^\lambda_a + \omega^\lambda_c \wedge \omega^c_a + \omega ^\lambda_\mu \wedge\omega^\mu_a)e_\lambda $$ $$ DDe_\lambda = (d\omega^a_\lambda + \omega^a_c \wedge \omega^c_\lambda + \omega ^a_\mu \wedge \omega^\mu_\lambda)e_a + (d\omega^\mu_\lambda + \omega^\mu_c \wedge \omega^c_\lambda + \omega ^\mu_\nu \wedge \omega^\nu_\lambda)e_\mu $$ Here one can easily identify the $S$ and $S^\perp$ curvatures: $$ R_Se_a = (d\omega^b_a + \omega^b_c \wedge \omega^c_a)e_b$$ $$ R_{S^\perp} e_\lambda = (d\omega^\mu_\lambda + \omega^\mu_c \wedge)e_\mu$$ and here the book states \begin{align*} R e_a = (R_S + B\wedge A + DA)e_a\\ R e_\lambda = (R_{S^\perp} + A\wedge B + DB)e_\lambda \end{align*} but I am certainly confused about how to calculate the action of either $A\wedge B$, $DA$ or $DB$ since I don't know which version of the Leibnitz rule to implement, namely: $$ (A\wedge B) e_a = A^\lambda_b \wedge B^c _\mu e_a$$ or inverting the order of the wedges.
Clarification: for the $A,B$ operators one can see that their matrix expression in the chosen split frame is $$Ae_a = \omega^\lambda_b e_\lambda~~ B e_\lambda = \omega^a_\lambda e_a$$