Holomorphic vector bundles on almost complex manifolds

As far as I understand, being Kähler only needs an almost complex manifold

This is false.

A Kähler manifold is a complex manifold equipped with a hermitian metric such that the associated two-form is closed. The corresponding notion for an almost complex manifold as opposed to a complex manifold is called an almost Kähler manifold.

Your confusion has little to do with the definition of Kähler, but rather with the definition of a holomorphic vector bundle.

By definition, a holomorphic vector bundle is a complex vector bundle over a complex manifold $X$ such that the total space $E$ is a complex manifold and the projection map $\pi : E \to X$ is holomorphic. If $X$ is not complex, no bundle $\pi : E \to X$ is holomorphic.

In particular, while an almost complex structure on $M$ gives rise to a splitting $TM\otimes\mathbb{C} = T^{1,0}M\oplus T^{0,1}M$, the complex bundle $T^{1,0}M \to M$ is a holomorphic vector bundle only when the almost complex structure is integrable, in which case $M$ is complex.

Note, calling $T^{0,1}M$ the antiholomorphic tangent bundle is potentially misleading as there is no such thing as an antiholomorphic vector bundle. A holomorphic vector bundle has holomorphic transition functions, but a vector bundle cannot have antiholomorphic transition functions as the composition of antiholomorphic functions is holomorphic.

Take two smooth manifolds $N$ and $M$ with almost-complex structures $I$ and $J$, respectively. Consider a smooth map $f \colon N \to M$, with derivative $f_* \colon TN \to TM$.

When $I$ and $J$ are complex structures, $f$ is holomorphic if and only if $J f_* = f_* I$. If we use this characterization as the definition of "holomorphic," we get a definition that makes sense even when $I$ and $J$ are merely almost-complex structures.

Now, consider a complex vector bundle $\pi \colon E \to M$. From the definition of a complex vector bundle, $E$ comes with a complex structure $i \colon E_m \to E_m$ on every fiber. Pick an almost-complex structure $I$ on the total space $E$ which is compatible with the fiber-wise complex structure, in the sense that it agrees with the action of $i$ on vertical tangent vectors. (A "vertical tangent vector" is a vector in the kernel of $\pi_* \colon TE \to TM$. I'll leave it as an exercise to guess how $i$ is supposed to act on vertical tangent vectors.)

Now we can say $E$ is "holomorphic" if $\pi$ is holomorphic, using our earlier definition of holomorphicity for maps between almost-complex manifolds.

This answer comes with the standard caveat that even if you can generalize a definition, it doesn't mean you should. This definition of a holomorphic vector bundle may only turn out to be useful when one or both of $I$ and $J$ are complex structures. At the very least, however, this definition should help you understand which results about holomorphic vector bundles fail, and why, when the base space or the total space is merely almost-complex.