Suppose $E \to M$ is a complex vector bundle over a complex manifold $M$. As I understand it, there are two possible ways to say what it means that $E$ is a holomorphic vector bundle:
$E$ has the structure of a complex manifold, such that the projection to $M$ is a holomorphic map. This means that there is an endomorphism $J:TE\to TE$ of the ($C^\infty$) tangent bundle of $E$ that squares to $-1$, and such that $J$ is integrable, in the sense that its Nijenhuis tensor vanishes (or, equivalently, the operator $\bar{\partial}:\Omega^0_E \to \Omega^{0,1}_E$ squares to $0$). We also require that $\pi$ respects $J$ in the sense that $d\pi$ commutes with the complex structures on $E$ and $M$.
$E$ has a bundle endomorphism $\tilde{J}:E\to E$ such that $\tilde{J}^2=-1$, subject to an integrability condition that the associated Dolbeault operator $\bar{\partial}_E: \Omega^0_M\otimes E \to \Omega^{0,1}\otimes E$ squares to $0$.
How do we see that these definitions are equivalent (if they even are)? More specifically, I'd like to see how one can move between the two notions of almost complex structure $J$ and $\tilde{J}$, as well as how to ensure the two integrability conditions match.
Here is a slightly different way to rephrase my question: there is a Newlander-Nirenberg theorem for integrable almost complex structures on the tangent bundle (which is what I use in definition 1), and there is a "linearized" version of Newlander-Nirenberg for vector bundles (like in definition 2). How does one relate the two theorems? Are they equivalent for vector bundles?
Edit: I have an idea on how to begin the relation. If we have a $\tilde{J}$ as in (2) above, we could try $J = d\tilde{J}: TE \to TE$. Since $\tilde{J}^2= - \text{id}$, the differential should satisfy $J^2 = -\text{id}$ by the chain rule.
It's not quite clear if integrability transfers over using this idea, or how the corresponding Dolbeault operators relate.