[proofplan]
The assertion is local, so we work in holomorphic coordinate charts on $X$ and $Y$. In such coordinates, a form of type $(p,q)$ is a sum of coefficient functions multiplied by wedges of exactly $p$ holomorphic coordinate differentials and exactly $q$ antiholomorphic coordinate differentials. Holomorphicity of $F$ implies that the pullback of each holomorphic coordinate differential on $Y$ is a linear combination only of holomorphic coordinate differentials on $X$, and similarly for antiholomorphic differentials after conjugation. Linearity and compatibility of pullback with wedge products then preserve the number of holomorphic and antiholomorphic factors in every local term.
[/proofplan]
[step:Express the form locally in holomorphic coordinates]
Let $x_0 \in X$ be arbitrary, and set $y_0 := F(x_0) \in Y$. Choose a holomorphic coordinate chart $(U,\varphi)$ on $X$ with $x_0 \in U$, where
\begin{align*}
\varphi: U \to \varphi(U) \subset \mathbb{C}^m
\end{align*}
has coordinate functions $z_1,\dots,z_m: U \to \mathbb{C}$. Choose a holomorphic coordinate chart $(V,\psi)$ on $Y$ with $y_0 \in V$ and, after replacing $U$ by the open neighbourhood $U \cap F^{-1}(V)$ of $x_0$, assume $F(U) \subset V$. Write
\begin{align*}
\psi: V \to \psi(V) \subset \mathbb{C}^n
\end{align*}
with coordinate functions $w_1,\dots,w_n: V \to \mathbb{C}$.
Since $\alpha \in A^{p,q}(Y)$, on $V$ it has the local form
\begin{align*}
\alpha|_V = \sum_{I,J} a_{I,J}\, dw_I \wedge d\bar w_J
\end{align*}
where $I=(i_1,\dots,i_p)$ ranges over strictly increasing $p$-tuples in $\{1,\dots,n\}$, $J=(j_1,\dots,j_q)$ ranges over strictly increasing $q$-tuples in $\{1,\dots,n\}$, each coefficient is a smooth function $a_{I,J}: V \to \mathbb{C}$, and
\begin{align*}
dw_I := dw_{i_1}\wedge \cdots \wedge dw_{i_p}
\end{align*}
while
\begin{align*}
d\bar w_J := d\bar w_{j_1}\wedge \cdots \wedge d\bar w_{j_q}.
\end{align*}
If $p=0$ or $q=0$, the corresponding wedge factor is interpreted as $1$.
[/step]
[step:Compute the pullback of holomorphic coordinate differentials]
For each $a \in \{1,\dots,n\}$, define the smooth coordinate component
\begin{align*}
f_a: U \to \mathbb{C}, \qquad f_a := w_a \circ F.
\end{align*}
Because $F$ is holomorphic and $w_a$ is a holomorphic coordinate function on $V$, the function $f_a$ is holomorphic on $U$. Hence, in the coordinates $z_1,\dots,z_m$,
\begin{align*}
\frac{\partial f_a}{\partial \bar z_r} = 0
\end{align*}
for every $r \in \{1,\dots,m\}$. The differential of a smooth complex-valued function in complex coordinates is
\begin{align*}
df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r + \sum_{r=1}^m \frac{\partial f_a}{\partial \bar z_r}\, d\bar z_r.
\end{align*}
Therefore
\begin{align*}
F^*(dw_a) = d(w_a \circ F) = df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r.
\end{align*}
Thus $F^*(dw_a)$ is a linear combination of the holomorphic coordinate one-forms $dz_1,\dots,dz_m$ and has type $(1,0)$.
[guided]
Fix $a \in \{1,\dots,n\}$. The coordinate function $w_a: V \to \mathbb{C}$ selects the $a$-th holomorphic coordinate on $Y$. Pulling this coordinate function back along $F$ gives the smooth function
\begin{align*}
f_a: U \to \mathbb{C}, \qquad f_a := w_a \circ F.
\end{align*}
Since $F$ is holomorphic and $w_a$ is holomorphic in the chosen chart on $Y$, the composite $f_a$ is holomorphic in the chosen chart on $X$. In complex coordinates, holomorphicity is exactly the vanishing of the antiholomorphic derivatives:
\begin{align*}
\frac{\partial f_a}{\partial \bar z_r} = 0
\end{align*}
for every $r \in \{1,\dots,m\}$.
Now compute the pullback of the coordinate one-form $dw_a$. Pullback commutes with exterior differentiation on functions, so
\begin{align*}
F^*(dw_a) = d(w_a \circ F) = df_a.
\end{align*}
For a smooth complex-valued function, the [exterior derivative](/theorems/1525) decomposes into holomorphic and antiholomorphic coordinate parts:
\begin{align*}
df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r + \sum_{r=1}^m \frac{\partial f_a}{\partial \bar z_r}\, d\bar z_r.
\end{align*}
The second sum vanishes because $f_a$ is holomorphic. Hence
\begin{align*}
F^*(dw_a) = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r.
\end{align*}
This is the key point: no $d\bar z_r$ term appears. Therefore the pullback of a holomorphic coordinate differential is again a form of type $(1,0)$ on $X$.
[/guided]
[/step]
[step:Compute the pullback of antiholomorphic coordinate differentials]
For each $a \in \{1,\dots,n\}$, the antiholomorphic coordinate function on $V$ is
\begin{align*}
\bar w_a: V \to \mathbb{C}, \qquad y \mapsto \overline{w_a(y)}.
\end{align*}
Using $f_a = w_a \circ F$ as above, we have
\begin{align*}
\bar w_a \circ F = \overline{f_a}.
\end{align*}
Since $f_a$ is holomorphic, $\overline{f_a}$ is antiholomorphic in the coordinates $z_1,\dots,z_m$, so
\begin{align*}
\frac{\partial \overline{f_a}}{\partial z_r} = 0
\end{align*}
for every $r \in \{1,\dots,m\}$. Therefore
\begin{align*}
F^*(d\bar w_a) = d(\bar w_a \circ F) = d\overline{f_a} = \sum_{r=1}^m \frac{\partial \overline{f_a}}{\partial \bar z_r}\, d\bar z_r.
\end{align*}
Thus $F^*(d\bar w_a)$ is a linear combination of the antiholomorphic coordinate one-forms $d\bar z_1,\dots,d\bar z_m$ and has type $(0,1)$.
[/step]
[step:Apply multiplicativity and linearity of pullback to each local summand]
Pullback is linear over sums, pulls back coefficient functions by composition, and is compatible with wedge products. Hence on $U$,
\begin{align*}
F^*(\alpha|_V) = \sum_{I,J} (a_{I,J}\circ F)\, F^*(dw_I) \wedge F^*(d\bar w_J).
\end{align*}
For $I=(i_1,\dots,i_p)$,
\begin{align*}
F^*(dw_I) = F^*(dw_{i_1})\wedge \cdots \wedge F^*(dw_{i_p}),
\end{align*}
and each factor $F^*(dw_{i_s})$ is a linear combination of $dz_1,\dots,dz_m$. Therefore $F^*(dw_I)$ is a sum of wedge products containing exactly $p$ holomorphic coordinate differentials. Similarly, for $J=(j_1,\dots,j_q)$,
\begin{align*}
F^*(d\bar w_J) = F^*(d\bar w_{j_1})\wedge \cdots \wedge F^*(d\bar w_{j_q}),
\end{align*}
and each factor is a linear combination of $d\bar z_1,\dots,d\bar z_m$, so $F^*(d\bar w_J)$ is a sum of wedge products containing exactly $q$ antiholomorphic coordinate differentials.
Multiplying by the smooth coefficient function $a_{I,J}\circ F: U \to \mathbb{C}$ does not change type. Hence every local summand of $F^*\alpha$ has type $(p,q)$ on $U$, and therefore
\begin{align*}
F^*\alpha|_U \in A^{p,q}(U).
\end{align*}
[/step]
[step:Conclude the type condition globally]
The point $x_0 \in X$ was arbitrary. For every point of $X$, we have found a holomorphic coordinate neighbourhood on which $F^*\alpha$ is represented as a smooth sum of terms with exactly $p$ holomorphic coordinate differentials and exactly $q$ antiholomorphic coordinate differentials. Since being of type $(p,q)$ is a local condition on a complex manifold, it follows that
\begin{align*}
F^*\alpha \in A^{p,q}(X).
\end{align*}
This proves that holomorphic pullback preserves type $(p,q)$.
[/step]
