$\textbf{Hint: }$ After you choose vectors $v_{k+1},\dots,v_n$, you have linearly independent vectors on $T_pM$. Extend $\{v_{k+1},\dots,v_n\}$ around a neighbourhood of $p$, say to constant local vector fields $X_{k+1},\dots,X_n$. Since $\{X_1|_p,\dots,X_k|_p,X_{k+1}|_p,\dots,X_n|_p\}$ linearly independent, the matrix $[X_i^j(p)]$ is invertible. Use continuity of determinant function to show that there is a smaller neighbourhood of $p$ such that $\{X_1,\dots,X_n\}$ is linearly independent there.
First the isomorphism between the complex bundle $(T_{\Bbb{R}}X,I)$ and $T^{(1,0)}X$ is really the linear algebra, pick the frame $$(T_{\Bbb{R}}X,I) \to T^{(1,0)}X\\\frac{\partial}{\partial x_i}\mapsto \frac{1}{2}(\frac{\partial}{\partial x_i } - i \frac{\partial}{\partial y_i})$$
If can be checked this is a complex isomorphism, as the local frame is isomorphic the vector bundle is also isomorphic.
For the first one and the third one :
Since $$\frac{\partial}{\partial z_i} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} -i\frac{\partial}{\partial y_i}\right)$$
We know in the real case $$\frac{\partial}{\partial x_j} = \frac{\partial \tilde{x}_j}{\partial x_i} \frac{\partial}{\partial \tilde{x_j}}$$
Similar for $y_i$,
Substitute into the above definition gives that $$\frac{\partial }{\partial z_i} = \frac{\partial \tilde{z}_j}{\partial z_i} \frac{\partial}{\partial \tilde{z_j}}$$
Therefore the cocycle is really given in the first definition,
therefore we see these three definitions are equivalent.
Maybe these three different characterizations have their own strength
for example the cocycle characterization will be useful in building some abstract theory about vector bundles (for example when dealing with classification problems).
The third one is very useful in local computation since we are familiar with differential calculus on the local coordinate for the real case, the calculation will naturally extend to the complexified case in the third setting, that's why this is useful.
Best Answer
You are right. If you want the statement is true around a neighborhood you should look at @Neal answer instead of mine in that post, which more or less refer to the following theorem.
So not all local frame expressible as coordinate frame. Only the commuting ones.