[proofplan]
The formula is pointwise, so we fix one point in the coordinate chart and work in the complex cotangent space there. The Hermitian coefficient matrix $H=(h_{i\bar j})$ is positive Hermitian, hence it can be written as $H=B^\top\overline{B}$ for an invertible complex matrix $B$; this matrix changes the coordinate coframe into a unitary coframe with the stated covariant metric convention. In that unitary coframe the volume form is the chosen normalizing constant times the standard real oriented volume form. The only remaining issue is to track how a complex coframe change acts on real top exterior forms, and the real determinant is $|\det B|^2=\det H$.
[/proofplan]
[step:Fix a point and replace the coordinate coframe by a unitary coframe]
Fix a point $p\in U$. Let $T_p^{1,0}X$ denote the holomorphic tangent space at $p$, and let $T_p^{*(1,0)}X$ denote its complex dual. The coordinate $(1,0)$-coframe at $p$ is
\begin{align*}
(dz_1)_p,\dots,(dz_n)_p\in T_p^{*(1,0)}X.
\end{align*}
The Hermitian metric has coefficient matrix
\begin{align*}
H(p)=(h_{i\bar j}(p))_{i,j=1}^n\in \mathbb{C}^{n\times n}.
\end{align*}
Since $h$ is Hermitian and positive definite, $H(p)$ is a positive Hermitian matrix.
By the finite-dimensional spectral theorem for positive Hermitian matrices, $H(p)$ has a positive Hermitian square root $S(p)=H(p)^{1/2}$. Define
\begin{align*}
B(p)=(B_{ki}(p))_{k,i=1}^n:=S(p)^\top\in GL(n,\mathbb{C}).
\end{align*}
Since $S(p)$ is Hermitian, $\overline{S(p)^\top}=S(p)$, and therefore
\begin{align*}
B(p)^\top\overline{B(p)}=S(p)\overline{S(p)^\top}=S(p)^2=H(p).
\end{align*} Define complex covectors
\begin{align*}
\alpha_k(p)=\sum_{i=1}^n B_{ki}(p)\,(dz_i)_p
\end{align*}
for $k\in\{1,\dots,n\}$. Then the coefficient of $(dz_i)_p\otimes (d\bar z_j)_p$ in $\sum_{k=1}^n \alpha_k(p)\otimes \overline{\alpha_k(p)}$ is $\sum_{k=1}^n B_{ki}(p)\overline{B_{kj}(p)}=(B(p)^\top\overline{B(p)})_{ij}=h_{i\bar j}(p)$. Hence
\begin{align*}
\sum_{k=1}^n \alpha_k(p)\otimes \overline{\alpha_k(p)}
=
\sum_{i,j=1}^n h_{i\bar j}(p)\,(dz_i)_p\otimes (d\bar z_j)_p.
\end{align*}
Thus $\alpha_1(p),\dots,\alpha_n(p)$ is a unitary coframe for $h_p$ in the covariant sense specified in the statement.
[guided]
We work at a single point $p\in U$ because the desired identity is an identity of top-degree forms, and such an identity can be checked pointwise. At $p$, the holomorphic coordinate covectors
\begin{align*}
(dz_1)_p,\dots,(dz_n)_p
\end{align*}
form a complex basis of the complex cotangent space $T_p^{*(1,0)}X$. In this basis, the Hermitian metric is represented by the matrix
\begin{align*}
H(p)=(h_{i\bar j}(p))_{i,j=1}^n.
\end{align*}
The hypothesis that $h$ is a Hermitian metric means precisely that $H(p)$ is Hermitian and positive definite.
The point of introducing a new coframe is that the volume form has its simplest expression in a unitary coframe. Since $H(p)$ is positive Hermitian, the finite-dimensional spectral theorem gives a positive Hermitian square root $S(p)=H(p)^{1/2}$. We choose
\begin{align*}
B(p)=(B_{ki}(p))_{k,i=1}^n:=S(p)^\top.
\end{align*}
Then $B(p)$ is invertible because $H(p)$ is positive definite. Also, since $S(p)$ is Hermitian, $\overline{S(p)^\top}=S(p)$, and hence
\begin{align*}
B(p)^\top\overline{B(p)}=S(p)\overline{S(p)^\top}=S(p)^2=H(p).
\end{align*}
Define the new complex covectors by
\begin{align*}
\alpha_k(p)=\sum_{i=1}^n B_{ki}(p)\,(dz_i)_p
\end{align*}
for each $k\in\{1,\dots,n\}$. Expanding the standard Hermitian form in the $\alpha$-coframe gives
\begin{align*}
\sum_{k=1}^n \alpha_k(p)\otimes \overline{\alpha_k(p)}
=
\sum_{k=1}^n\sum_{i,j=1}^n B_{ki}(p)\overline{B_{kj}(p)}\,(dz_i)_p\otimes(d\bar z_j)_p.
\end{align*}
The coefficient of $(dz_i)_p\otimes(d\bar z_j)_p$ is
\begin{align*}
\sum_{k=1}^n B_{ki}(p)\overline{B_{kj}(p)}=(B(p)^\top\overline{B(p)})_{ij}.
\end{align*}
By the chosen factorization $H(p)=B(p)^\top\overline{B(p)}$, this coefficient is $h_{i\bar j}(p)$. Therefore
\begin{align*}
\sum_{k=1}^n \alpha_k(p)\otimes \overline{\alpha_k(p)}=h_p.
\end{align*}
This is exactly the unitary-coframe condition specified in the theorem statement, so $\alpha_1(p),\dots,\alpha_n(p)$ is a unitary coframe for $h_p$.
[/guided]
[/step]
[step:Compute the normalized volume form in the unitary coframe]
For each $k\in\{1,\dots,n\}$, write
\begin{align*}
\alpha_k(p)=a_k(p)+i b_k(p)
\end{align*}
with $a_k(p)$ and $b_k(p)$ real covectors on the real cotangent space $T_p^*X$. By the pointwise normalization of the Hermitian volume form stated in the theorem, applied to the coframe $\alpha_1(p),\dots,\alpha_n(p)$ satisfying $h_p=\sum_{k=1}^n \alpha_k(p)\otimes\overline{\alpha_k(p)}$, the value of $dV_h$ at $p$ is
\begin{align*}
(dV_h)_p=C_n\, a_1(p)\wedge b_1(p)\wedge\cdots\wedge a_n(p)\wedge b_n(p).
\end{align*}
Here $C_n>0$ depends only on the convention used to pass from the fundamental form to the volume form, and not on $p$ or on the coordinate chart.
[/step]
[step:Track the real determinant of the complex coframe change]
Let $A(p)$ be the real $2n\times 2n$ matrix describing the real coframe change from
\begin{align*}
(dx_1)_p,(dy_1)_p,\dots,(dx_n)_p,(dy_n)_p
\end{align*}
to
\begin{align*}
a_1(p),b_1(p),\dots,a_n(p),b_n(p).
\end{align*}
Since $\alpha_k(p)=\sum_{i=1}^n B_{ki}(p)\,(dz_i)_p$ for each $k$, the real matrix $A(p)$ is the realification of the complex matrix $B(p)$. If $B=C+iD$, with $C,D\in\mathbb{R}^{n\times n}$, this realification is the real [linear map](/page/Linear%20Map) on $\mathbb{R}^n\times\mathbb{R}^n$ sending $(u,v)$ to $(Cu-Dv,Du+Cv)$. Its real determinant is
\begin{align*}
\det_{\mathbb{R}} A(p)=|\det_{\mathbb{C}} B(p)|^2.
\end{align*}
Therefore the top exterior product transforms as
\begin{align*}
a_1(p)\wedge b_1(p)\wedge\cdots\wedge a_n(p)\wedge b_n(p)
=
|\det_{\mathbb{C}}B(p)|^2\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
[guided]
We now convert the complex change of coframe into a statement about real oriented volume forms. This is the step where the square of the complex determinant appears.
Write the complex matrix $B(p)$ as
\begin{align*}
B(p)=C(p)+iD(p),
\end{align*}
where $C(p),D(p)\in\mathbb{R}^{n\times n}$. The component relations
\begin{align*}
\alpha_k(p)=\sum_{i=1}^n B_{ki}(p)(dz_i)_p
\end{align*}
for $k\in\{1,\dots,n\}$ mean that, after writing $\alpha_k(p)=a_k(p)+i b_k(p)$ and $(dz_j)_p=(dx_j)_p+i(dy_j)_p$, the real covectors $a_k(p)$ and $b_k(p)$ are obtained from $(dx_j)_p$ and $(dy_j)_p$ by the real linear map on $\mathbb{R}^n\times\mathbb{R}^n$ sending $(u,v)$ to $(C(p)u-D(p)v,D(p)u+C(p)v)$. This is the standard realification of the complex linear map $B(p):\mathbb{C}^n\to\mathbb{C}^n$.
The standard realification determinant formula for complex linear maps says that a complex matrix $B(p)\in GL(n,\mathbb{C})$, viewed as a real linear map on $\mathbb{R}^{2n}$, has real determinant equal to the squared modulus of its complex determinant. Applying this formula to $B(p)$ gives
\begin{align*}
\det_{\mathbb{R}} A(p)=|\det_{\mathbb{C}}B(p)|^2.
\end{align*}
The exterior product of a real coframe is multiplied by the determinant of the real change-of-coframe matrix. Hence
\begin{align*}
a_1(p)\wedge b_1(p)\wedge\cdots\wedge a_n(p)\wedge b_n(p)
=
|\det_{\mathbb{C}}B(p)|^2\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
The sign is positive because $|\det_{\mathbb{C}}B(p)|^2>0$, so the complex-linear change preserves the natural real orientation determined by the ordered coframe $(dx_1,dy_1,\dots,dx_n,dy_n)$.
[/guided]
[/step]
[step:Identify the determinant factor with $\det(h_{i\bar j})$]
Taking determinants in the matrix identity $H(p)=B(p)^\top\overline{B(p)}$ gives
\begin{align*}
\det H(p)=\det(B(p)^\top)\det(\overline{B(p)}).
\end{align*}
Since $\det(B(p)^\top)=\det B(p)$ and $\det(\overline{B(p)})=\overline{\det B(p)}$, this becomes
\begin{align*}
\det H(p)=|\det B(p)|^2.
\end{align*}
Substituting this identity into the unitary-coframe volume formula yields
\begin{align*}
(dV_h)_p
=
C_n\det H(p)\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
Since $H(p)=(h_{i\bar j}(p))$, this is
\begin{align*}
(dV_h)_p
=
C_n\det(h_{i\bar j}(p))\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
The point $p\in U$ was arbitrary, so the identity holds on the whole coordinate chart:
\begin{align*}
dV_h=C_n\det(h_{i\bar j})\, dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n.
\end{align*}
This is the claimed coordinate expression for the Hermitian volume form.
[guided]
The determinant factor in the real volume change must now be rewritten in terms of the metric coefficients. The matrix identity already proved is
\begin{align*}
H(p)=B(p)^\top\overline{B(p)}.
\end{align*}
Taking determinants and using multiplicativity of the determinant gives
\begin{align*}
\det H(p)=\det(B(p)^\top)\det(\overline{B(p)}).
\end{align*}
The determinant is invariant under transpose, so $\det(B(p)^\top)=\det B(p)$. Complex conjugation commutes with the determinant, so $\det(\overline{B(p)})=\overline{\det B(p)}$. Hence
\begin{align*}
\det H(p)=\det B(p)\,\overline{\det B(p)}=|\det B(p)|^2.
\end{align*}
Substituting this into the unitary-coframe formula and the realification determinant formula gives
\begin{align*}
(dV_h)_p=C_n\det H(p)\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
Since $H(p)=(h_{i\bar j}(p))_{i,j=1}^n$, this is
\begin{align*}
(dV_h)_p=C_n\det(h_{i\bar j}(p))\,(dx_1)_p\wedge(dy_1)_p\wedge\cdots\wedge(dx_n)_p\wedge(dy_n)_p.
\end{align*}
Because $p\in U$ was arbitrary, the pointwise identity is exactly the form identity
\begin{align*}
dV_h=C_n\det(h_{i\bar j})\, dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n
\end{align*}
on the coordinate chart $U$.
[/guided]
[/step]
