Let $X$ be a complex manifold of dimension $n$ (holomorphic charts). As a real smooth manifold, it has dimension $2n$. I am trying to think what is exactly the tangent bundle of $X$. If we think of tangent vectors as $\mathbb{C}$-linear derivation of smooth complex valued functions, then by the answer in this stackexchange post Complexified tangent space
$$Der (C^\infty_{X,p,\mathbb C}\to \mathbb C)=T_{\mathbb{R},p}(X)\otimes_\mathbb{R}\mathbb{C}= Der_\mathbb R (X^\infty_{X,p,\mathbb R}\to \mathbb R) \otimes_\mathbb R \mathbb C $$
Therefore as a complex vector space, the tangent space of $X$ has dimension $2n$ (unlike the real case,dimension of tangent space of complex manifold is 2 times the dimension of the complex manifold), and in fact it has local decomposition of $v = v^i \partial_{z_i} + v^{\bar{j}} \partial_{\bar{z_j}}$ where $v^i,v^\bar{j}$ are smooth complex valued functions on $X$.
Now as noted in the answer above, we did not use the fact that the transition maps are holomorphic at all. In fact, we could start with a real smooth manifold of dimension $2n$ to achieve the same thing. So by Newlander-Nierenberg theorem, if we can give the real manifold an integrable almost complex structure $J$, then we will able to find holomorphic charts for the manifold, thus making the real manifold into a complex one.
Please let me know if my understanding is incorrect or not.
Best Answer
I'm not entirely sure what your question is, but you have one misconception. You confused the complex tangent space with the complexification of the real tangent space.
The complex tangent space is a subspace of $T_{\mathbb C,p} X$, defined by $$T^{1,0}_p X= \{v \in T_{\mathbb C,p}X : (I\otimes \mathbb C)(v) = iv\}.$$ In other words, it is the eigenspace of $I \otimes \mathbb C$ to the eigenvalue $i$. The set of vectors where application of $I$ is the same as multiplication by $i$.
If $z_1, \dotsc, z_n$ are local holomorphic coordinates around $p$, then $T^{1,0}_pX$ is generated by the derivations $\frac \partial {\partial z_1}, \dotsc, \frac \partial {\partial z_n}$. So it has complex dimension $n$.
The connection to holomorphic maps comes from the following observation:
Exercise. A smooth map $f: X \to Y$ between complex manifolds is holomorphic at $p \in X$ if and only if the complexified differential map $T_pf \otimes \mathbb C: T_{\mathbb C,p}X \to T_{\mathbb C,f(p)}Y$ maps $T^{1,0}_pX$ to $T^{1,0}_{f(p)}Y$.
Also note that as complex smooth bundles, $T_{\mathbb R} X$ and $T^{1,0}X$ are isomorphic: We can split $T_{\mathbb C}X = T^{1,0}X \oplus T^{0,1}X$ where $T^{0,1} = \{v : (I \otimes \mathbb C) = -iv\}, $and then the composition $T_{\mathbb R}X \hookrightarrow T_{\mathbb C}X \twoheadrightarrow T^{1,0}X$ is an isomorphism of smooth $\mathbb C$-bundles.