[proofplan]
We compare the two pullbacks of the global curvature form $\Omega$ along the local sections $s_i$ and $s_j$. Since $s_j(x)=s_i(x)g_{ij}(x)$, differentiating $s_j$ produces one contribution from the differential of $s_i$ and one vertical contribution from the differential of $g_{ij}$. The curvature form is horizontal, so all terms involving the vertical contribution vanish. The remaining term is transformed by the $G$-equivariance of curvature under the right action, giving exactly the adjoint transformation law.
[/proofplan]
[step:Rewrite the second section through the right action and differentiate it]
Let
\begin{align*}
U_{ij}:=U_i\cap U_j.
\end{align*}
For each $h \in G$, let
\begin{align*}
R_h:P\to P
\end{align*}
denote the right action map $p\mapsto ph$. Define the smooth map
\begin{align*}
\Phi:U_{ij}\to P,\quad x\mapsto s_i(x)g_{ij}(x).
\end{align*}
By the definition of the transition function, $\Phi=s_j|_{U_{ij}}$.
Fix $x\in U_{ij}$ and $v\in T_xM$. Choose a smooth curve
\begin{align*}
\gamma:(-\varepsilon,\varepsilon)\to U_{ij}
\end{align*}
with $\gamma(0)=x$ and $\gamma'(0)=v$. Define
\begin{align*}
p:(-\varepsilon,\varepsilon)\to P,\quad t\mapsto s_i(\gamma(t))
\end{align*}
and
\begin{align*}
h:(-\varepsilon,\varepsilon)\to G,\quad t\mapsto g_{ij}(\gamma(t)).
\end{align*}
Then
\begin{align*}
s_j(\gamma(t))=p(t)h(t).
\end{align*}
Writing $h_0:=h(0)=g_{ij}(x)$, differentiation of the action map gives
\begin{align*}
(ds_j)_x(v)=d(R_{h_0})_{s_i(x)}\bigl((ds_i)_x(v)\bigr)+\xi_v^\#{}_{s_j(x)},
\end{align*}
where $\xi_v\in\mathfrak g$ is the Lie algebra element represented by the tangent vector $h'(0)\in T_{h_0}G$ after right translation to the identity, and $\xi_v^\#$ denotes the corresponding fundamental vertical vector field on $P$.
[guided]
We first make precise what changes when the section is multiplied by the transition function. Set $U_{ij}:=U_i\cap U_j$. For each $h\in G$, the right action map is
\begin{align*}
R_h:P\to P,\quad p\mapsto ph.
\end{align*}
The transition relation says that the map
\begin{align*}
\Phi:U_{ij}\to P,\quad x\mapsto s_i(x)g_{ij}(x)
\end{align*}
is exactly the restricted section $s_j|_{U_{ij}}$.
Now fix $x\in U_{ij}$ and a tangent vector $v\in T_xM$. To compute $(ds_j)_x(v)$, represent $v$ by a smooth curve
\begin{align*}
\gamma:(-\varepsilon,\varepsilon)\to U_{ij}
\end{align*}
with $\gamma(0)=x$ and $\gamma'(0)=v$. Along this curve, define
\begin{align*}
p:(-\varepsilon,\varepsilon)\to P,\quad t\mapsto s_i(\gamma(t))
\end{align*}
and
\begin{align*}
h:(-\varepsilon,\varepsilon)\to G,\quad t\mapsto g_{ij}(\gamma(t)).
\end{align*}
Then the curve whose derivative we need is $t\mapsto p(t)h(t)$. Put $h_0:=h(0)=g_{ij}(x)$. Differentiating the action map separates the variation of $p(t)$ from the variation of $h(t)$:
\begin{align*}
(ds_j)_x(v)=d(R_{h_0})_{s_i(x)}\bigl((ds_i)_x(v)\bigr)+\xi_v^\#{}_{s_j(x)}.
\end{align*}
Here $\xi_v\in\mathfrak g$ is obtained from $h'(0)\in T_{h_0}G$ by right translation back to $T_eG=\mathfrak g$, and $\xi_v^\#$ is the fundamental vertical vector field generated by $\xi_v$. The important point is the type of the two summands: the first is the transported tangent vector coming from $s_i$, while the second is vertical because it comes only from moving in the principal $G$-fiber.
[/guided]
[/step]
[step:Use horizontality to discard every vertical contribution]
Fix $x\in U_{ij}$ and $v,w\in T_xM$. From the previous step, there exist $\xi_v,\xi_w\in\mathfrak g$ such that
\begin{align*}
(ds_j)_x(v)=d(R_{g_{ij}(x)})_{s_i(x)}\bigl((ds_i)_x(v)\bigr)+\xi_v^\#{}_{s_j(x)}
\end{align*}
and
\begin{align*}
(ds_j)_x(w)=d(R_{g_{ij}(x)})_{s_i(x)}\bigl((ds_i)_x(w)\bigr)+\xi_w^\#{}_{s_j(x)}.
\end{align*}
The curvature form $\Omega$ is horizontal, so $\Omega_p(a,b)=0$ whenever at least one of $a,b\in T_pP$ is vertical. Since each fundamental vector $\xi^\#_p$ is vertical, bilinearity of $\Omega_{s_j(x)}$ gives
\begin{align*}
\Omega_{s_j(x)}\bigl((ds_j)_x(v),(ds_j)_x(w)\bigr)=\Omega_{s_j(x)}\bigl(d(R_{g_{ij}(x)})_{s_i(x)}(ds_i)_x(v),d(R_{g_{ij}(x)})_{s_i(x)}(ds_i)_x(w)\bigr).
\end{align*}
[/step]
[step:Apply equivariance of the global curvature form]
The curvature form of a principal connection is equivariant under the right action:
\begin{align*}
(R_h)^*\Omega=\operatorname{Ad}_{h^{-1}}\Omega
\end{align*}
for every $h\in G$. Applying this identity with $h=g_{ij}(x)$, $p=s_i(x)$, $a=(ds_i)_x(v)$, and $b=(ds_i)_x(w)$ gives
\begin{align*}
\Omega_{s_j(x)}\bigl(d(R_{g_{ij}(x)})_{s_i(x)}a,d(R_{g_{ij}(x)})_{s_i(x)}b\bigr)=\operatorname{Ad}_{g_{ij}(x)^{-1}}\bigl(\Omega_{s_i(x)}(a,b)\bigr).
\end{align*}
Substituting $a=(ds_i)_x(v)$ and $b=(ds_i)_x(w)$, and using the preceding step, we obtain
\begin{align*}
\Omega_{s_j(x)}\bigl((ds_j)_x(v),(ds_j)_x(w)\bigr)=\operatorname{Ad}_{g_{ij}(x)^{-1}}\bigl(\Omega_{s_i(x)}((ds_i)_x(v),(ds_i)_x(w))\bigr).
\end{align*}
[/step]
[step:Identify both sides as the local curvature forms]
By the definition of pullback of a $\mathfrak g$-valued $2$-form,
\begin{align*}
(F_j)_x(v,w)=\Omega_{s_j(x)}\bigl((ds_j)_x(v),(ds_j)_x(w)\bigr)
\end{align*}
and
\begin{align*}
(F_i)_x(v,w)=\Omega_{s_i(x)}\bigl((ds_i)_x(v),(ds_i)_x(w)\bigr).
\end{align*}
Therefore, for every $x\in U_{ij}$ and every $v,w\in T_xM$,
\begin{align*}
(F_j)_x(v,w)=\operatorname{Ad}_{g_{ij}(x)^{-1}}\bigl((F_i)_x(v,w)\bigr).
\end{align*}
This is precisely the identity
\begin{align*}
F_j=\operatorname{Ad}_{g_{ij}^{-1}}F_i
\end{align*}
on $U_i\cap U_j$.
[/step]
