I've found out that the primary motivation for complexifying the tangent bundle is to enable its decomposition into the eigenspaces $\pm i$ of $J$. This decomposition facilitates the splitting of the bundle into holomorphic and antiholomorphic components. For the complexified tangent space $T_pM^\mathbb C$, we have a complex basis given by $\{\partial/\partial x_1,\dots, \partial/\partial x_n, \partial/\partial y_1,\dots, \partial/\partial y_n\}$. Given that $T_pM^\mathbb C \cong T_pM^{1,0} \oplus T_pM^{0,1}$, the isomorphisms can be represented as: \begin{align*} T_pM &\to T_pM^{1,0}, \ v \mapsto \frac{1}{2}(v-iJv) \\ T_pM &\to T_pM^{0,1}, \ v \mapsto \frac{1}{2}(v+iJv). \end{align*}
Next I wanted to show that under these isomorphisms we can map the basis $$\{\partial/\partial x_1,\dots, \partial/\partial x_n, \partial/\partial x_y,\dots, \partial/\partial y_n\}$$ to the basis $$\{\partial/\partial z_1,\dots, \partial/\partial z_n, \partial/\partial \bar{z}_1,\dots, \partial/\partial \bar{z}_n\}$$ but I get some redundancy. For the first map $T_pM \to T_pM^{1,0}, \ v \mapsto \frac{1}{2}(v-iJv)$ we have \begin{align*} \frac{\partial}{\partial x_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial x_j} - iJ\left(\frac{\partial}{\partial x_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j}\right) = \frac{\partial}{\partial z_j} \\ \frac{\partial}{\partial y_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial y_j} - iJ\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} + i\frac{\partial}{\partial x_j}\right) \end{align*}
and here $\frac{\partial}{\partial y_j}$ doesn't map to either $\frac{\partial}{\partial z_j}$ or $\frac{\partial}{\partial \bar{z}_j}$.
Similarly for the map $T_pM \to T_pM^{0,1}, \ v \mapsto \frac{1}{2}(v+iJv)$ we have
\begin{align*} \frac{\partial}{\partial x_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial x_j} + iJ\left(\frac{\partial}{\partial x_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial x_j} + i\frac{\partial}{\partial y_j}\right) = \frac{\partial}{\partial \bar{z}_j} \\ \frac{\partial}{\partial y_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial y_j} + iJ\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} - i\frac{\partial}{\partial x_j}\right) \end{align*}
but again $\frac{\partial}{\partial y_j}$ doesn't map to a corresponding basis element. Could anyone help me out why?
Edit: Does the following work out properly with the differences regarding complex and real bases? Since $T_pM^\mathbb C \cong T_pM^{1,0} \oplus T_pM^{0,1}$ we can use the isomorphisms \begin{align*} T_pM &\to T_pM^{1,0}, \ v \mapsto \frac{1}{2}(v-iJv) \\ T_pM &\to T_pM^{0,1}, \ v \mapsto \frac{1}{2}(v+iJv) \end{align*} to send the real basis $\left\{ \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}, \frac{\partial}{\partial y_1}, \dots, \frac{\partial}{\partial y_n} \right\}$ of $T_pM$ to \begin{align*} \frac{\partial}{\partial x_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial x_j} - iJ\left(\frac{\partial}{\partial x_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j}\right) = \frac{\partial}{\partial z_j} \\ \frac{\partial}{\partial y_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial y_j} - iJ\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} + i\frac{\partial}{\partial x_j}\right) = i\frac{\partial}{\partial z_j} \end{align*}
giving $T_pM^{1,0}$ a real basis $\left\{ \frac{\partial}{\partial z_1}, \dots, \frac{\partial}{\partial z_n}, i\frac{\partial}{\partial z_1}, \dots, i\frac{\partial}{\partial z_n} \right\}$ and a complex basis $\left\{ \frac{\partial}{\partial z_1}, \dots, \frac{\partial}{\partial z_n}\right\}$. In a similar manner we send the real basis of $T_pM$ to $T_pM^{0,1}$ by \begin{align*} \frac{\partial}{\partial x_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial x_j} + iJ\left(\frac{\partial}{\partial x_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial x_j} + i\frac{\partial}{\partial y_j}\right) = \frac{\partial}{\partial \bar{z}_j} \\ \frac{\partial}{\partial y_j} &\mapsto \frac{1}{2}\left(\frac{\partial}{\partial y_j} + iJ\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} - i\frac{\partial}{\partial x_j}\right) = -i\frac{\partial}{\partial \bar{z}_j} \end{align*}
which gives $T_pM^{0,1}$ a real basis $\left\{ \frac{\partial}{\partial z_1}, \dots, \frac{\partial}{\partial z_n}, -i\frac{\partial}{\partial \bar{z}_1}, \dots, -i\frac{\partial}{\partial \bar{z}_n} \right\}$ and a complex basis $\left\{ \frac{\partial}{\partial \bar{z}_1}, \dots, \frac{\partial}{\partial \bar{z}_n}\right\}$.
The formulae you have derive show that under the first map $\frac{\partial}{\partial y_j}$ maps to $i\frac{\partial}{\partial z_j}$ while under the second map it maps to $-i\frac{\partial}{\partial \bar z_j}$, which is consistent with the fact that the map to $T^{1,0}M$ is complex linear while the map to $T^{0,1}M$ is conjugate linear. You also have to keep in mind that you start from a real basis of $T_p M$ while the $ \frac{\partial}{\partial z_j}$ form a complex basis for $T^{1,0}M$. (And by definition, the vectors $\frac{\partial}{\partial X_j}$ do form a $\mathbb C$-basis for $T_pM$, so they should be mapped to a $\mathbb C$-basis of $T^{1,0}M$.)