[proofplan]
We compute both sides on the real coordinate frame associated to the holomorphic chart. First we express the real metric coefficients of $g$ in terms of the Hermitian coefficients $h_{i\bar j}$ using the convention $g=2\operatorname{Re}h$. Then the identities $J\partial_{x_i}=\partial_{y_i}$ and $J\partial_{y_i}=-\partial_{x_i}$ give the real coordinate coefficients of $\omega$. Finally we expand the proposed complex-coordinate expression and verify that it has exactly the same values on the real frame.
[/proofplan]
[step:Express the real coordinate frame in terms of the holomorphic frame]
Fix the coordinate chart $(U,z)$ and write $z_i=x_i+i y_i$. For each $i\in\{1,\dots,n\}$, define the real coordinate vector fields
\begin{align*}
X_i=\partial_{x_i}
\end{align*}
and
\begin{align*}
Y_i=\partial_{y_i}.
\end{align*}
The associated holomorphic coordinate vector field is
\begin{align*}
\partial_{z_i}=\frac{1}{2}(X_i-iY_i).
\end{align*}
The $(1,0)$-parts of the real vector fields are therefore
\begin{align*}
(X_i)^{1,0}=\partial_{z_i}
\end{align*}
and
\begin{align*}
(Y_i)^{1,0}=i\partial_{z_i}.
\end{align*}
The complex structure acts on the real coordinate frame by
\begin{align*}
JX_i=Y_i
\end{align*}
and
\begin{align*}
JY_i=-X_i.
\end{align*}
[/step]
[step:Compute the real metric coefficients from the Hermitian coefficients]
For $i,j\in\{1,\dots,n\}$, set
\begin{align*}
h_{i\bar j}=h(\partial_{z_i},\partial_{z_j}).
\end{align*}
Since $h$ is Hermitian with the stated convention, $h_{j\bar i}=\overline{h_{i\bar j}}$. Using $g(u,v)=2\operatorname{Re}h(u^{1,0},v^{1,0})$, we obtain
\begin{align*}
g(X_i,X_j)=2\operatorname{Re}h_{i\bar j}.
\end{align*}
Because $h$ is conjugate-linear in its second argument,
\begin{align*}
g(X_i,Y_j)=2\operatorname{Re}h(\partial_{z_i},i\partial_{z_j})=2\operatorname{Re}(-i h_{i\bar j})=2\operatorname{Im}h_{i\bar j}.
\end{align*}
Similarly, because $h$ is linear in its first argument,
\begin{align*}
g(Y_i,X_j)=2\operatorname{Re}h(i\partial_{z_i},\partial_{z_j})=2\operatorname{Re}(i h_{i\bar j})=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
Finally,
\begin{align*}
g(Y_i,Y_j)=2\operatorname{Re}h(i\partial_{z_i},i\partial_{z_j})=2\operatorname{Re}h_{i\bar j}.
\end{align*}
[guided]
The goal of this step is to remove any ambiguity about the factor of $2$ and the signs. The real metric is not obtained by simply taking the real part of $h$ on the complex frame; it is obtained from the $(1,0)$-parts of real tangent vectors by the normalization
\begin{align*}
g(u,v)=2\operatorname{Re}h(u^{1,0},v^{1,0}).
\end{align*}
For the real coordinate vector fields, the $(1,0)$-parts are
\begin{align*}
(X_i)^{1,0}=\partial_{z_i}
\end{align*}
and
\begin{align*}
(Y_i)^{1,0}=i\partial_{z_i}.
\end{align*}
Thus
\begin{align*}
g(X_i,X_j)=2\operatorname{Re}h(\partial_{z_i},\partial_{z_j})=2\operatorname{Re}h_{i\bar j}.
\end{align*}
The mixed term is where the convention on complex linearity matters. Since $h$ is conjugate-linear in the second argument, scalar multiplication by $i$ in the second slot produces the factor $\overline{i}=-i$. Hence
\begin{align*}
h(\partial_{z_i},i\partial_{z_j})=-i h(\partial_{z_i},\partial_{z_j})=-i h_{i\bar j}.
\end{align*}
Taking twice the real part gives
\begin{align*}
g(X_i,Y_j)=2\operatorname{Re}(-i h_{i\bar j})=2\operatorname{Im}h_{i\bar j}.
\end{align*}
In the reversed mixed term, the scalar $i$ lies in the first argument, where $h$ is complex-linear. Therefore
\begin{align*}
h(i\partial_{z_i},\partial_{z_j})=i h_{i\bar j}.
\end{align*}
Taking twice the real part gives
\begin{align*}
g(Y_i,X_j)=2\operatorname{Re}(i h_{i\bar j})=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
Finally, the two factors of $i$ cancel because the first slot contributes $i$ and the second slot contributes $\overline{i}=-i$:
\begin{align*}
h(i\partial_{z_i},i\partial_{z_j})=i(-i)h_{i\bar j}=h_{i\bar j}.
\end{align*}
Thus
\begin{align*}
g(Y_i,Y_j)=2\operatorname{Re}h_{i\bar j}.
\end{align*}
[/guided]
[/step]
[step:Compute the fundamental form on the real coordinate frame]
By definition,
\begin{align*}
\omega(u,v)=g(Ju,v)
\end{align*}
for real tangent vectors $u,v$. Using $JX_i=Y_i$ and $JY_i=-X_i$, the metric coefficients computed above give
\begin{align*}
\omega(X_i,X_j)=g(Y_i,X_j)=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
Also,
\begin{align*}
\omega(X_i,Y_j)=g(Y_i,Y_j)=2\operatorname{Re}h_{i\bar j}.
\end{align*}
Similarly,
\begin{align*}
\omega(Y_i,X_j)=g(-X_i,X_j)=-2\operatorname{Re}h_{i\bar j}.
\end{align*}
Finally,
\begin{align*}
\omega(Y_i,Y_j)=g(-X_i,Y_j)=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
[/step]
[step:Expand the proposed complex formula and compare coefficients]
Define the real $2$-form $\alpha$ on $U$ by
\begin{align*}
\alpha=i\sum_{p,q=1}^{n}h_{p\bar q}\,dz_p\wedge d\bar z_q.
\end{align*}
Since $dz_p(X_i)=\delta_{pi}$, $dz_p(Y_i)=i\delta_{pi}$, $d\bar z_q(X_j)=\delta_{qj}$, and $d\bar z_q(Y_j)=-i\delta_{qj}$, direct evaluation gives
\begin{align*}
\alpha(X_i,X_j)=i(h_{i\bar j}-h_{j\bar i})=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
Next,
\begin{align*}
\alpha(X_i,Y_j)=h_{i\bar j}+h_{j\bar i}=2\operatorname{Re}h_{i\bar j}.
\end{align*}
Also,
\begin{align*}
\alpha(Y_i,X_j)=-(h_{i\bar j}+h_{j\bar i})=-2\operatorname{Re}h_{i\bar j}.
\end{align*}
Finally,
\begin{align*}
\alpha(Y_i,Y_j)=i(h_{i\bar j}-h_{j\bar i})=-2\operatorname{Im}h_{i\bar j}.
\end{align*}
These four identities agree exactly with the corresponding values of $\omega$ on the frame vectors $X_i,Y_i$.
[/step]
[step:Conclude equality of the forms]
The real vector fields $X_1,Y_1,\dots,X_n,Y_n$ form a local frame of the real tangent bundle $TU$ over the coordinate chart. Since $\omega$ and $\alpha$ are real bilinear alternating $2$-forms and they agree on every pair of frame vectors, they agree on all real tangent vectors on $U$. Therefore
\begin{align*}
\omega=\alpha=i\sum_{i,j=1}^{n}h_{i\bar j}\,dz_i\wedge d\bar z_j.
\end{align*}
The same coefficient comparison also shows that the displayed complex expression is real as a real differential form, because all of its values on the real frame are [real numbers](/page/Real%20Numbers). This proves the claimed local formula.
[/step]
