[proofplan]
We compare the same local section written in the two holomorphic frames. The metric transformation follows by substituting the coordinate relation $a=g\tilde a$ into the Hermitian norm formula. The connection transformation follows by applying the definition of the local connection matrix to $s=e g\tilde a=\tilde e\tilde a$ and comparing coefficients in the frame $e$. Finally, the curvature transformation follows from applying $D^2$ to the same section and using the tensorial defining formula for the curvature matrix.
[/proofplan]
[step:Relate the coefficient vectors in the two frames]
Let
\begin{align*}
a:U&\to\mathbb{C}^r,\\
\tilde a:U&\to\mathbb{C}^r
\end{align*}
be smooth coefficient maps for the same smooth local section $s:U\to E$. Since $e$ is a frame and $\tilde e=e g$, the identity
\begin{align*}
s=\tilde e\tilde a=e a
\end{align*}
is equivalent to
\begin{align*}
e g\tilde a=e a.
\end{align*}
Because $e_1,\dots,e_r$ are pointwise linearly independent, this gives the coefficient relation
\begin{align*}
a=g\tilde a.
\end{align*}
[guided]
The frame $e=(e_1,\dots,e_r)$ is being used as a row vector of sections. Thus multiplying it by a column vector $a:U\to\mathbb{C}^r$ gives the section
\begin{align*}
e a=\sum_{i=1}^r e_i a_i.
\end{align*}
The changed frame is defined by $\tilde e=e g$, where $g:U\to GL(r,\mathbb{C})$ is a holomorphic matrix-valued map. If the same section is written as $s=e a=\tilde e\tilde a$, then
\begin{align*}
e a=\tilde e\tilde a=e g\tilde a.
\end{align*}
Since $e_1,\dots,e_r$ form a basis of each fiber $E_x$, equality of sections written in the frame $e$ is exactly equality of their coefficient vectors. Hence
\begin{align*}
a=g\tilde a.
\end{align*}
This relation is the only coordinate change needed in the rest of the proof.
[/guided]
[/step]
[step:Substitute the coefficient relation into the Hermitian norm]
For every smooth coefficient map $\tilde a:U\to\mathbb{C}^r$, the section $s=\tilde e\tilde a=e(g\tilde a)$ satisfies
\begin{align*}
|s|_h^2
&=\overline{g\tilde a}^{\top}h(g\tilde a)\\
&=\overline{\tilde a}^{\top}\overline{g}^{\top}h g\tilde a\\
&=\overline{\tilde a}^{\top}g^*h g\tilde a.
\end{align*}
By definition of the metric matrix in the frame $\tilde e$, the same norm is
\begin{align*}
|s|_h^2=\overline{\tilde a}^{\top}\tilde h\tilde a.
\end{align*}
Since this identity holds for all vectors $\tilde a(x)\in\mathbb{C}^r$ at each point $x\in U$, the Hermitian matrices agree pointwise:
\begin{align*}
\tilde h=g^*hg.
\end{align*}
[/step]
[step:Compare the $(1,0)$ connection term in the two frames]
Let $\tilde a:U\to\mathbb{C}^r$ be a smooth coefficient map and set $s=\tilde e\tilde a=e(g\tilde a)$. Using the definition of $A$ in the frame $e$ and the product rule for the Dolbeault operator $\partial$, we obtain
\begin{align*}
D^{1,0}s
&=D^{1,0}(e(g\tilde a))\\
&=e\bigl(\partial(g\tilde a)+A g\tilde a\bigr)\\
&=e\bigl((\partial g)\tilde a+g\partial\tilde a+A g\tilde a\bigr).
\end{align*}
Using instead the definition of $\tilde A$ in the frame $\tilde e=e g$ gives
\begin{align*}
D^{1,0}s
&=D^{1,0}(\tilde e\tilde a)\\
&=\tilde e(\partial\tilde a+\tilde A\tilde a)\\
&=e g(\partial\tilde a+\tilde A\tilde a)\\
&=e\bigl(g\partial\tilde a+g\tilde A\tilde a\bigr).
\end{align*}
Comparing the coefficient vectors in the frame $e$ yields
\begin{align*}
(\partial g)\tilde a+g\partial\tilde a+A g\tilde a
=
g\partial\tilde a+g\tilde A\tilde a.
\end{align*}
Canceling $g\partial\tilde a$ gives
\begin{align*}
\bigl((\partial g)+A g-g\tilde A\bigr)\tilde a=0.
\end{align*}
Since $\tilde a(x)\in\mathbb{C}^r$ is arbitrary at each point $x\in U$, we have
\begin{align*}
g\tilde A=(\partial g)+Ag.
\end{align*}
Multiplying on the left by $g^{-1}$ gives
\begin{align*}
\tilde A=g^{-1}Ag+g^{-1}\partial g.
\end{align*}
[guided]
The connection matrix is defined by how the $(1,0)$ part of the Chern connection acts on a section written in a chosen frame. We compute $D^{1,0}s$ twice for the same section
\begin{align*}
s=\tilde e\tilde a=e(g\tilde a),
\end{align*}
where $\tilde a:U\to\mathbb{C}^r$ is an arbitrary smooth coefficient map.
First use the frame $e$. Since $a=g\tilde a$, the defining identity for $A$ gives
\begin{align*}
D^{1,0}s
=
D^{1,0}(e(g\tilde a))
=
e\bigl(\partial(g\tilde a)+A g\tilde a\bigr).
\end{align*}
The Dolbeault operator $\partial$ satisfies the matrix product rule, so
\begin{align*}
\partial(g\tilde a)=(\partial g)\tilde a+g\partial\tilde a.
\end{align*}
Therefore
\begin{align*}
D^{1,0}s
=
e\bigl((\partial g)\tilde a+g\partial\tilde a+A g\tilde a\bigr).
\end{align*}
Now compute the same object in the frame $\tilde e$. By definition of $\tilde A$,
\begin{align*}
D^{1,0}s
=
D^{1,0}(\tilde e\tilde a)
=
\tilde e(\partial\tilde a+\tilde A\tilde a).
\end{align*}
Since $\tilde e=e g$, this becomes
\begin{align*}
D^{1,0}s
=
e g(\partial\tilde a+\tilde A\tilde a)
=
e\bigl(g\partial\tilde a+g\tilde A\tilde a\bigr).
\end{align*}
Both formulas are expansions of the same section-valued $(1,0)$-form in the same frame $e$. Hence their coefficient vectors agree:
\begin{align*}
(\partial g)\tilde a+g\partial\tilde a+A g\tilde a
=
g\partial\tilde a+g\tilde A\tilde a.
\end{align*}
Canceling the common term $g\partial\tilde a$ gives
\begin{align*}
\bigl((\partial g)+A g-g\tilde A\bigr)\tilde a=0.
\end{align*}
Because $\tilde a$ was arbitrary, at each point this matrix-valued $(1,0)$-form must vanish:
\begin{align*}
(\partial g)+A g-g\tilde A=0.
\end{align*}
Since $g(x)\in GL(r,\mathbb{C})$ for every $x\in U$, left multiplication by $g^{-1}$ is valid and gives
\begin{align*}
\tilde A=g^{-1}Ag+g^{-1}\partial g.
\end{align*}
[/guided]
[/step]
[step:Apply $D^2$ to the same section to transform the curvature]
Let $\tilde a:U\to\mathbb{C}^r$ be a smooth coefficient map and set $s=\tilde e\tilde a=e(g\tilde a)$. By the definition of the curvature matrix $\Theta$ in the frame $e$,
\begin{align*}
D^2s
=
D^2(e(g\tilde a))
=
e(\Theta g\tilde a).
\end{align*}
By the definition of the curvature matrix $\tilde\Theta$ in the frame $\tilde e$,
\begin{align*}
D^2s
=
D^2(\tilde e\tilde a)
=
\tilde e(\tilde\Theta\tilde a)
=
e g\tilde\Theta\tilde a.
\end{align*}
Comparing coefficients in the frame $e$ gives
\begin{align*}
\Theta g\tilde a=g\tilde\Theta\tilde a.
\end{align*}
Since $\tilde a(x)\in\mathbb{C}^r$ is arbitrary at each point $x\in U$,
\begin{align*}
\Theta g=g\tilde\Theta.
\end{align*}
Multiplying on the left by $g^{-1}$ yields
\begin{align*}
\tilde\Theta=g^{-1}\Theta g.
\end{align*}
Together with the metric and connection transformations proved above, this proves all three gauge transformation laws.
[/step]
