I follow Kobayashi "Differential Geometry of Complex Vector Bundles", pages 11-12, prop. 4.9. Given a rank-$r$ Hermitian holomorphic vector bundle $(E,h)$ over a complex manifold $M$, there exists a unique $h$-connection $D$ such that $D^{0,1}=\bar\partial$.

The proof works in a local holomorphic frame $s=(s_1,...,s_r)$ and uses uniqueness to construct as follows: Consider the connection form $\omega^j_i$: $$0=\bar\partial s_i=D^{0,1}(s_i)=s_j{\omega^{0,1}}^j_i $$ so $\omega^j_i$ is of type $(1,0)$.

Since $D$ is an $h$-connection, $$dh_{ij}=h(Ds_i,s_j)+h(s_i,Ds_j)=\omega_i^ah_{a\bar j}+h_{ib}\bar \omega^b_{\bar j}. $$

Now the claim is that since $\omega$ is $(1,0)$ then $$d'h_{ij}=\omega^a_ih_{a\bar j}$$.

Now here is my confusion: This is (as good as I understand) supposed to mean that (in local coordinates on M) $\bar \omega(\partial_{z^l})$=0. But this conjugation of $\omega$ is on the target, i.e. conjugate the **values** of $\omega$ and not the domain, i.e. $\partial_{z^l}$.

In this case, it is unlikely that $\bar \omega(\partial_{z^l})=0$ since this would mean that $\omega(\partial_{z^l})=0$ so $\omega$ is both type $(1,0)$ and $(0,1)$, that is: zero.

The question comes down to this: There are two ways to conjugate a form - conjugate the input, and conjugate the output. In this calculation, I understand that we use the latter kind, but the conclusion that the type is exchanged assumes the former kind.