[proofplan]
The pullback identity reduces the gauge-transformed local form to the connection form evaluated in the changed section $s'=su$. We compute the derivative of $s'$ at a point $x\in U$ and split it into a part obtained by right-translating $ds_x(X)$ and a vertical fundamental vector generated by $(u^{-1}du)_x(X)$. The right-equivariance of a principal connection evaluates the first part as $\operatorname{Ad}_{u(x)^{-1}}A_x(X)$, while the reproduction property evaluates the vertical part as $(u^{-1}du)_x(X)$. Since this holds for every tangent vector $X\in T_xU$, the asserted one-form identity follows.
[/proofplan]
[step:Reduce the gauge-transformed pullback to the changed section]
Define the changed local section $s':U\to P$ by
\begin{align*}
s'(x)=s(x)u(x).
\end{align*}
By the representing property of $u$, we have $\Phi\circ s=s'$. Therefore, using functoriality of pullback,
\begin{align*}
A'=s^*(\Phi^*\omega)=(\Phi\circ s)^*\omega=(s')^*\omega.
\end{align*}
Thus it is enough to compute $(s')^*\omega$.
[/step]
[step:Split the derivative of the changed section into translated and vertical parts]
Fix $x\in U$ and $X\in T_xU$. Let $\gamma:(-\varepsilon,\varepsilon)\to U$ be a smooth curve satisfying $\gamma(0)=x$ and $\gamma'(0)=X$. Define
\begin{align*}
p:=s(x)\in P,\qquad g:=u(x)\in G.
\end{align*}
Also define the curve $h:(-\varepsilon,\varepsilon)\to G$ by
\begin{align*}
h(t)=u(x)^{-1}u(\gamma(t)).
\end{align*}
Let $e_G\in G$ denote the identity element of the Lie group $G$. Then $h(0)=e_G$, and its velocity at $0$, identified with an element of $\mathfrak g=T_{e_G}G$, is
\begin{align*}
h'(0)=(u^{-1}du)_x(X).
\end{align*}
Let $\exp:\mathfrak g\to G$ denote the Lie group exponential map. For each $t$,
\begin{align*}
s'(\gamma(t))=s(\gamma(t))u(\gamma(t))=\bigl(s(\gamma(t))g\bigr)h(t).
\end{align*}
Let $\rho:P\times G\to P$ denote the principal right action, so that $\rho(q,a)=qa$. Differentiating the identity $s'(\gamma(t))=\rho(s(\gamma(t))g,h(t))$ at $t=0$ and using the differential $d\rho_{(pg,e_G)}:T_{pg}P\times T_{e_G}G\to T_{pg}P$ gives
\begin{align*}
d s'_x(X)=d(R_g)_p(ds_x(X))+\bigl((u^{-1}du)_x(X)\bigr)_P(pg),
\end{align*}
because the first component is the derivative of fixed right multiplication by $g$ and the second component is the fundamental vertical vector generated by $h'(0)$. Here $R_g:P\to P$ is right multiplication by $g$, and for $\xi\in\mathfrak g$ the fundamental vertical vector field $\xi_P$ is defined by
\begin{align*}
\xi_P(q)=\left.\frac{d}{dt}\right|_{t=0}q\exp(t\xi)
\end{align*}
for $q\in P$.
[guided]
We want to understand the tangent vector $ds'_x(X)$, but the section $s'$ changes in two ways as $x$ moves: the point $s(x)$ in the principal bundle moves, and the group element $u(x)$ multiplying it also moves. The purpose of this step is to separate those two contributions.
Choose a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to U$ with $\gamma(0)=x$ and $\gamma'(0)=X$. Define
\begin{align*}
p:=s(x)\in P,\qquad g:=u(x)\in G.
\end{align*}
The curve whose derivative we need is
\begin{align*}
t\mapsto s'(\gamma(t))=s(\gamma(t))u(\gamma(t)).
\end{align*}
To isolate the variation of $u$, define $h:(-\varepsilon,\varepsilon)\to G$ by
\begin{align*}
h(t)=u(x)^{-1}u(\gamma(t)).
\end{align*}
Let $e_G\in G$ denote the identity element of $G$. Then $h(0)=e_G$. Its tangent vector at the identity is precisely the left Maurer-Cartan derivative of $u$ applied to $X$:
\begin{align*}
h'(0)=(u^{-1}du)_x(X)\in\mathfrak g.
\end{align*}
Let $\exp:\mathfrak g\to G$ denote the Lie group exponential map.
Now rewrite the curve $s'(\gamma(t))$ as
\begin{align*}
s'(\gamma(t))=\bigl(s(\gamma(t))g\bigr)h(t).
\end{align*}
The first factor $s(\gamma(t))g$ records the motion of $s(\gamma(t))$ transported by fixed right multiplication $R_g:P\to P$. Its derivative at $0$ is
\begin{align*}
d(R_g)_p(ds_x(X)).
\end{align*}
The second factor $h(t)$ starts at the identity and therefore contributes the vertical fundamental vector at $pg=s(x)u(x)$ generated by $h'(0)=(u^{-1}du)_x(X)$. By the definition of the fundamental vertical vector field,
\begin{align*}
\bigl((u^{-1}du)_x(X)\bigr)_P(pg)=\left.\frac{d}{dt}\right|_{t=0}pg\exp\bigl(t(u^{-1}du)_x(X)\bigr).
\end{align*}
Let $\rho:P\times G\to P$ denote the principal right action, $\rho(q,a)=qa$. Differentiating $s'(\gamma(t))=\rho(s(\gamma(t))g,h(t))$ at $t=0$ uses the map $d\rho_{(pg,e_G)}:T_{pg}P\times T_{e_G}G\to T_{pg}P$. The first input vector is $d(R_g)_p(ds_x(X))$, because $g$ is fixed in the curve $s(\gamma(t))g$. The second input vector is $h'(0)=(u^{-1}du)_x(X)$, and by the definition of the fundamental vertical vector field it contributes $\bigl((u^{-1}du)_x(X)\bigr)_P(pg)$. Thus
\begin{align*}
d s'_x(X)=d(R_g)_p(ds_x(X))+\bigl((u^{-1}du)_x(X)\bigr)_P(pg).
\end{align*}
[/guided]
[/step]
[step:Evaluate the translated part using equivariance of the connection form]
By the defining right-equivariance property in the definition of a principal connection,
\begin{align*}
(R_g)^*\omega=\operatorname{Ad}_{g^{-1}}\omega
\end{align*}
for every $g\in G$. Applying this identity at $p=s(x)$ to the tangent vector $ds_x(X)\in T_pP$ gives
\begin{align*}
\omega_{pg}\bigl(d(R_g)_p(ds_x(X))\bigr)=\operatorname{Ad}_{g^{-1}}\bigl(\omega_p(ds_x(X))\bigr).
\end{align*}
Since $g=u(x)$ and $A=s^*\omega$, this becomes
\begin{align*}
\omega_{s(x)u(x)}\bigl(d(R_{u(x)})_{s(x)}(ds_x(X))\bigr)=\operatorname{Ad}_{u(x)^{-1}}\bigl(A_x(X)\bigr).
\end{align*}
[/step]
[step:Evaluate the vertical part using the reproduction property]
By the defining reproduction property in the definition of a principal connection on fundamental vertical vector fields,
\begin{align*}
\omega_q(\xi_P(q))=\xi
\end{align*}
for every $q\in P$ and every $\xi\in\mathfrak g$. Applying this with
\begin{align*}
q=s(x)u(x),\qquad \xi=(u^{-1}du)_x(X),
\end{align*}
we obtain
\begin{align*}
\omega_{s(x)u(x)}\bigl(\bigl((u^{-1}du)_x(X)\bigr)_P(s(x)u(x))\bigr)=(u^{-1}du)_x(X).
\end{align*}
[/step]
[step:Combine the two evaluations to obtain the local transformation law]
Using the derivative splitting and the linearity of the one-form $\omega_{s(x)u(x)}:T_{s(x)u(x)}P\to\mathfrak g$, we compute
\begin{align*}
A'_x(X)=((s')^*\omega)_x(X)=\omega_{s(x)u(x)}(ds'_x(X)).
\end{align*}
Substituting the two pieces from the preceding steps gives
\begin{align*}
A'_x(X)=\operatorname{Ad}_{u(x)^{-1}}\bigl(A_x(X)\bigr)+(u^{-1}du)_x(X).
\end{align*}
Since $x\in U$ and $X\in T_xU$ were arbitrary, this proves the identity of $\mathfrak g$-valued one-forms
\begin{align*}
A'=\operatorname{Ad}_{u^{-1}}A+u^{-1}du.
\end{align*}
If $G$ is a matrix Lie group, the adjoint action is conjugation:
\begin{align*}
\operatorname{Ad}_{u(x)^{-1}}(\xi)=u(x)^{-1}\xi u(x)
\end{align*}
for every $x\in U$ and every $\xi\in\mathfrak g$. Therefore the same formula is written pointwise as
\begin{align*}
A'=u^{-1}Au+u^{-1}du.
\end{align*}
This completes the proof.
[/step]
