I am following pages 72 and 73 in Griffiths and Harris. Let $X$ be a complex manifold, and let $E \to X$ be a complex vector bundle. Then a connection $D$ is uniquely (locally) determined by a frame $\{e_i\}$ and a connection 1-form $\theta_{e}$. Let $\{f_i\}$ be another frame. Say the two frames are related by $e_i = \sum_j g_{ij} f_j$.Then the connection 1-form $\theta_{f}$ is given
$$ \theta_f = dg g^{-1} + g \theta_e g^{-1}. $$
We can decompose the complex cotangent bundle as $T^*X = T^{*\prime}X \oplus T^{*\prime \prime}X$, which are the holomorphic and antiholomorphic pieces, respectively. Thus we can write any connection $D$ as $D = D^{\prime} + D^{\prime \prime}$ where $D^{\prime}: \mathcal{A}^0(E) \to \mathcal{A}^{(1,0)}(E)$ and $D^{\prime \prime}: \mathcal{A}^0(E) \to \mathcal{A}^{(0,1)}(E)$. Locally, the connection 1-form $\theta$ is just a matrix of 1-forms. So $D^{\prime}$ would look like
$$ D^{\prime}\vert_U = (\partial \vert + \theta^{(1,0)}) \vert_U $$
and similarly for $D^{\prime \prime}$. Your $\theta^{(1,0)}$, for example, would be a matrix of $(1,0)$-forms. My question is, wouldn't the change of frame (change of gauge) change what types of forms you have in your connection 1-form? So how could you ever have a decomposition $D = D^{\prime} + D^{\prime \prime}$? If you change the frame then you would generically pick up some forms of every type from the $dg$ term for the new connection 1-form.
So is the decomposition really dependent on the frame? So then if I say, $\nabla$ is a connection of type $(1,0)$, am I implicitly assuming a choice of frame? And maybe with another frame if is not a connection of type $(1,0)$?