[proofplan]
We compare the two local sections on the overlap by differentiating the identity $s_j=s_i g_{ij}$. The derivative splits into a right-translated piece coming from $s_i$ and a vertical fundamental-vector-field piece coming from $g_{ij}$; applying the defining properties of a connection form gives the inhomogeneous transformation law for $A_j$. For curvature, we compute from the local formula $F=dA+\frac{1}{2}[A\wedge A]$ and use the Maurer-Cartan identity for $g_{ij}^{-1}dg_{ij}$ to cancel exactly the inhomogeneous terms. The remaining term is the adjoint transform of $F_i$.
[/proofplan]
[step:Differentiate the relation between the two local sections]
Set $W:=U_i\cap U_j$ and write $g:=g_{ij}:W\to G$. Let $\theta_L\in\Omega^1(G;\mathfrak g)$ be the left Maurer-Cartan form, and define
\begin{align*}
\alpha:=g^*\theta_L\in\Omega^1(W;\mathfrak g).
\end{align*}
Thus $\alpha=g^{-1}dg$.
Fix $x\in W$ and $v\in T_xW$. Choose a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to W$ with $\gamma(0)=x$ and $\gamma'(0)=v$. Put $p:=s_i(x)$ and $h:=g(x)$. The curve defining $s_j$ is
\begin{align*}
t\mapsto s_j(\gamma(t))=s_i(\gamma(t))g(\gamma(t)).
\end{align*}
Its tangent vector at $t=0$ decomposes as
\begin{align*}
(ds_j)_x(v)=(dR_h)_p((ds_i)_x(v))+(\alpha_x(v))^\#_{ph},
\end{align*}
where $R_h:P\to P$ is right multiplication by $h$, and $(\alpha_x(v))^\#_{ph}\in T_{ph}P$ is the fundamental vertical vector at $ph$ generated by $\alpha_x(v)\in\mathfrak g$.
Indeed, write
\begin{align*}
s_i(\gamma(t))g(\gamma(t))=\bigl(s_i(\gamma(t))h\bigr)\bigl(h^{-1}g(\gamma(t))\bigr).
\end{align*}
The first factor differentiates to $(dR_h)_p((ds_i)_x(v))$, while the second factor is the vertical curve through $ph$ whose velocity is generated by
\begin{align*}
\frac{d}{dt}\bigg|_{t=0}h^{-1}g(\gamma(t))=\theta_L{}_{g(x)}(dg_x(v))=\alpha_x(v).
\end{align*}
[/step]
[step:Apply the connection form to obtain the transformation law for $A_j$]
The defining properties of the connection form are
\begin{align*}
(R_h)^*\omega=\operatorname{Ad}_{h^{-1}}\omega
\end{align*}
for every $h\in G$, and
\begin{align*}
\omega_q(\xi^\#_q)=\xi
\end{align*}
for every $q\in P$ and every $\xi\in\mathfrak g$. Applying $\omega_{ph}$ to the tangent decomposition from the previous step gives
\begin{align*}
A_j{}_x(v)=\omega_{s_j(x)}((ds_j)_x(v)).
\end{align*}
Hence
\begin{align*}
A_j{}_x(v)=\operatorname{Ad}_{g(x)^{-1}}\bigl(\omega_{s_i(x)}((ds_i)_x(v))\bigr)+\alpha_x(v).
\end{align*}
Since $A_i=s_i^*\omega$, this is
\begin{align*}
A_j{}_x(v)=\operatorname{Ad}_{g(x)^{-1}}A_i{}_x(v)+\alpha_x(v).
\end{align*}
As $x\in W$ and $v\in T_xW$ were arbitrary,
\begin{align*}
A_j=\operatorname{Ad}_{g^{-1}}A_i+\alpha.
\end{align*}
Substituting $\alpha=g^{-1}dg$ gives
\begin{align*}
A_j=\operatorname{Ad}_{g^{-1}}A_i+g^{-1}dg.
\end{align*}
[guided]
We want to compute $A_j=s_j^*\omega$, so the only input is the derivative of the section $s_j$. On the overlap $W=U_i\cap U_j$, the relation between the sections is
\begin{align*}
s_j(x)=s_i(x)g(x),
\end{align*}
where $g=g_{ij}:W\to G$. The derivative has two sources: motion of the point $s_i(x)$ inside $P$, and motion of the group element $g(x)$ inside $G$.
Fix $x\in W$ and $v\in T_xW$. Choose a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to W$ with $\gamma(0)=x$ and $\gamma'(0)=v$. Let $p=s_i(x)$ and $h=g(x)$. Then
\begin{align*}
s_j(\gamma(t))=s_i(\gamma(t))g(\gamma(t)).
\end{align*}
To separate the two motions, rewrite this curve as
\begin{align*}
s_i(\gamma(t))g(\gamma(t))=\bigl(s_i(\gamma(t))h\bigr)\bigl(h^{-1}g(\gamma(t))\bigr).
\end{align*}
The derivative of $t\mapsto s_i(\gamma(t))h$ at $0$ is the right translate
\begin{align*}
(dR_h)_p((ds_i)_x(v)).
\end{align*}
The remaining factor $t\mapsto h^{-1}g(\gamma(t))$ is a curve in $G$ through the identity element. Its tangent vector is measured by the left Maurer-Cartan form. Since $\alpha:=g^*\theta_L$, this tangent is $\alpha_x(v)$. Therefore the second contribution to the tangent vector in $P$ is the fundamental vertical vector
\begin{align*}
(\alpha_x(v))^\#_{ph}.
\end{align*}
Thus
\begin{align*}
(ds_j)_x(v)=(dR_h)_p((ds_i)_x(v))+(\alpha_x(v))^\#_{ph}.
\end{align*}
Now apply the connection form $\omega$. The first defining property of a connection form says that right translation transforms $\omega$ by the adjoint action:
\begin{align*}
(R_h)^*\omega=\operatorname{Ad}_{h^{-1}}\omega.
\end{align*}
Hence
\begin{align*}
\omega_{ph}\bigl((dR_h)_p((ds_i)_x(v))\bigr)=\operatorname{Ad}_{h^{-1}}\bigl(\omega_p((ds_i)_x(v))\bigr).
\end{align*}
The second defining property says that $\omega$ returns the generator of a fundamental vertical vector:
\begin{align*}
\omega_{ph}\bigl((\alpha_x(v))^\#_{ph}\bigr)=\alpha_x(v).
\end{align*}
Adding the two contributions gives
\begin{align*}
A_j{}_x(v)=\operatorname{Ad}_{g(x)^{-1}}A_i{}_x(v)+\alpha_x(v).
\end{align*}
Because this equality holds for every $x\in W$ and every $v\in T_xW$, it is an equality of $\mathfrak g$-valued one-forms:
\begin{align*}
A_j=\operatorname{Ad}_{g^{-1}}A_i+\alpha.
\end{align*}
Finally, $\alpha=g^*\theta_L$ is precisely the standard notation $g^{-1}dg$, so
\begin{align*}
A_j=\operatorname{Ad}_{g^{-1}}A_i+g^{-1}dg.
\end{align*}
[/guided]
[/step]
[step:Record the differential identities used by the curvature computation]
For $\mathfrak g$-valued forms $\beta\in\Omega^k(W;\mathfrak g)$ and $\gamma\in\Omega^\ell(W;\mathfrak g)$, the bracket wedge product is the form $[\beta\wedge\gamma]\in\Omega^{k+\ell}(W;\mathfrak g)$ obtained by wedging the scalar form parts and applying the Lie bracket in $\mathfrak g$. In particular, for one-forms $\beta,\gamma\in\Omega^1(W;\mathfrak g)$,
\begin{align*}
[\beta\wedge\gamma]=[\gamma\wedge\beta].
\end{align*}
Define
\begin{align*}
B:=\operatorname{Ad}_{g^{-1}}A_i\in\Omega^1(W;\mathfrak g).
\end{align*}
Then
\begin{align*}
dB=\operatorname{Ad}_{g^{-1}}(dA_i)-[\alpha\wedge B].
\end{align*}
This follows by differentiating the pointwise adjoint action. In a matrix realization, it is the identity
\begin{align*}
d(g^{-1}A_i g)=g^{-1}(dA_i)g-(g^{-1}dg)\wedge(g^{-1}A_i g)-(g^{-1}A_i g)\wedge(g^{-1}dg),
\end{align*}
which is exactly the displayed formula because $[\alpha\wedge B]=\alpha\wedge B+B\wedge\alpha$ for one-forms.
The Maurer-Cartan identity for $\alpha=g^*\theta_L$ is
\begin{align*}
d\alpha+\frac{1}{2}[\alpha\wedge\alpha]=0.
\end{align*}
Equivalently, in a matrix realization, $d(g^{-1}dg)+(g^{-1}dg)\wedge(g^{-1}dg)=0$.
[/step]
[step:Compute the transformed curvature and cancel the inhomogeneous terms]
By the preceding transformation law for the local connection form,
\begin{align*}
A_j=B+\alpha.
\end{align*}
Using the local curvature formula for a connection form,
\begin{align*}
F_j=dA_j+\frac{1}{2}[A_j\wedge A_j].
\end{align*}
Substitute $A_j=B+\alpha$:
\begin{align*}
F_j=dB+d\alpha+\frac{1}{2}[B\wedge B]+[B\wedge\alpha]+\frac{1}{2}[\alpha\wedge\alpha].
\end{align*}
Using $dB=\operatorname{Ad}_{g^{-1}}(dA_i)-[\alpha\wedge B]$ and $[B\wedge\alpha]=[\alpha\wedge B]$ for one-forms, the mixed terms cancel:
\begin{align*}
F_j=\operatorname{Ad}_{g^{-1}}(dA_i)+\frac{1}{2}[B\wedge B]+d\alpha+\frac{1}{2}[\alpha\wedge\alpha].
\end{align*}
Using the Maurer-Cartan identity for $\alpha$, the last two terms vanish:
\begin{align*}
F_j=\operatorname{Ad}_{g^{-1}}(dA_i)+\frac{1}{2}[B\wedge B].
\end{align*}
Since $B=\operatorname{Ad}_{g^{-1}}A_i$ and $\operatorname{Ad}_{g^{-1}}:\mathfrak g\to\mathfrak g$ preserves the Lie bracket,
\begin{align*}
[B\wedge B]=\operatorname{Ad}_{g^{-1}}[A_i\wedge A_i].
\end{align*}
Therefore
\begin{align*}
F_j=\operatorname{Ad}_{g^{-1}}\left(dA_i+\frac{1}{2}[A_i\wedge A_i]\right).
\end{align*}
The expression in parentheses is $F_i$, so
\begin{align*}
F_j=\operatorname{Ad}_{g^{-1}}F_i.
\end{align*}
Restoring $g=g_{ij}$ gives the desired identity
\begin{align*}
F_j=\operatorname{Ad}_{g_{ij}^{-1}}F_i
\end{align*}
on $U_i\cap U_j$.
[/step]
