[proofplan] We compute the pullback $A_j=s_j^*\omega$ on an arbitrary tangent vector $X\in T_xU_{ij}$. Since $s_j$ is obtained from $s_i$ by the point-dependent right action of $g_{ij}$, the derivative of $s_j$ splits into a horizontal-looking variation of $s_i$ transported by right multiplication and a vertical fundamental vector field coming from the variation of $g_{ij}$. The connection form evaluates on these two terms by its defining equivariance and reproduction properties, giving respectively $\operatorname{Ad}(g_{ij}^{-1})A_i$ and $g_{ij}^*\theta^L$. [/proofplan] [step:Differentiate the section relation into horizontal and vertical contributions] Set $U:=U_{ij}$ and $g:=g_{ij}|_U:U\to G$. Let $\rho:P\times G\to P$ be the smooth right action map, $\rho(q,h)=q\cdot h$, and for each $h\in G$ let $R_h:P\to P$ be right multiplication, $R_h(q)=q\cdot h$. Fix $x\in U$ and $X\in T_xU$. Choose a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to U$ such that $\gamma(0)=x$ and $\gamma'(0)=X$. Define smooth curves $q:(-\varepsilon,\varepsilon)\to P$ and $h:(-\varepsilon,\varepsilon)\to G$ by $q(t):=s_i(\gamma(t))$ and $h(t):=g(\gamma(t))$. Then $s_j(\gamma(t))=q(t)\cdot h(t)$. Let $p:=s_j(x)=q(0)\cdot h(0)$ and define \begin{align*} \eta_X:=\theta^L_{g(x)}(dg_x(X))=(dL_{g(x)^{-1}})_{g(x)}(dg_x(X))\in\mathfrak g. \end{align*} The derivative of $t\mapsto q(t)\cdot h(t)$ at $t=0$ is \begin{align*} (ds_j)_x(X)=(dR_{g(x)})_{s_i(x)}((ds_i)_x(X))+(\eta_X)_P(p), \end{align*} where $(\eta_X)_P$ is the fundamental vector field on $P$ generated by $\eta_X$, namely \begin{align*} (\eta_X)_P(r)=\frac{d}{dt}\bigg|_{t=0} r\cdot \exp(t\eta_X) \end{align*} for $r\in P$. [guided] We need a precise formula for the derivative of the map $x\mapsto s_i(x)\cdot g(x)$. The difficulty is that both the point $s_i(x)\in P$ and the group element $g(x)\in G$ vary with $x$. Fix $x\in U$ and $X\in T_xU$. Choose a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to U$ with $\gamma(0)=x$ and $\gamma'(0)=X$. Define \begin{align*} q(t):=s_i(\gamma(t)) \end{align*} and \begin{align*} h(t):=g(\gamma(t)). \end{align*} Thus $q:(-\varepsilon,\varepsilon)\to P$ is a curve in the bundle and $h:(-\varepsilon,\varepsilon)\to G$ is a curve in the structure group. The section relation becomes \begin{align*} s_j(\gamma(t))=q(t)\cdot h(t). \end{align*} Set $p:=s_j(x)=q(0)\cdot h(0)$. The variation of $q(t)\cdot h(t)$ is the differential of the action map $\rho:P\times G\to P$ at $(s_i(x),g(x))$ applied to the tangent vector $((ds_i)_x(X),dg_x(X))$. By linearity of this differential, we split that tangent vector into $((ds_i)_x(X),0)+(0,dg_x(X))$. The first partial differential varies $q(t)$ while freezing the group element at $h(0)=g(x)$; this gives \begin{align*} (dR_{g(x)})_{s_i(x)}((ds_i)_x(X)). \end{align*} For the second partial differential, freeze the point $q(0)$ and vary the group element $h(t)$. To express this second piece as a fundamental vector field at $p$, translate the tangent vector $h'(0)=dg_x(X)\in T_{g(x)}G$ back to the Lie algebra by the left Maurer-Cartan form: \begin{align*} \eta_X:=\theta^L_{g(x)}(dg_x(X))=(dL_{g(x)^{-1}})_{g(x)}(dg_x(X))\in\mathfrak g. \end{align*} This choice is forced by the identity $g(x)^{-1}h(t)$ in matrix notation: the curve $g(x)^{-1}h(t)$ starts at the identity and has initial tangent $\eta_X$. Indeed, \begin{align*} q(0)\cdot h(t)=p\cdot\bigl(g(x)^{-1}h(t)\bigr). \end{align*} Let $\exp:\mathfrak g\to G$ denote the Lie group exponential map. By definition, the fundamental vector field generated by $\eta_X$ is the vector field $(\eta_X)_P:P\to TP$ given by \begin{align*} (\eta_X)_P(r)=\frac{d}{dt}\bigg|_{t=0} r\cdot \exp(t\eta_X) \end{align*} for $r\in P$. Therefore the derivative of this second curve at $t=0$ is the fundamental vector at $p$ generated by $\eta_X$: \begin{align*} \frac{d}{dt}\bigg|_{t=0} q(0)\cdot h(t)=(\eta_X)_P(p). \end{align*} Adding the two first-order contributions gives \begin{align*} (ds_j)_x(X)=(dR_{g(x)})_{s_i(x)}((ds_i)_x(X))+(\eta_X)_P(p). \end{align*} [/guided] [/step] [step:Evaluate the connection form on the transported section term] The connection form $\omega$ satisfies right-equivariance: \begin{align*} (R_a)^*\omega=\operatorname{Ad}(a^{-1})\omega \end{align*} for every $a\in G$. Applying this identity with $a=g(x)$ to the vector $(ds_i)_x(X)\in T_{s_i(x)}P$ gives \begin{align*} \omega_p\left((dR_{g(x)})_{s_i(x)}((ds_i)_x(X))\right)=\operatorname{Ad}(g(x)^{-1})\left(\omega_{s_i(x)}((ds_i)_x(X))\right). \end{align*} Since $A_i=s_i^*\omega$, this is \begin{align*} \omega_p\left((dR_{g(x)})_{s_i(x)}((ds_i)_x(X))\right)=\operatorname{Ad}(g(x)^{-1})(A_i)_x(X). \end{align*} [/step] [step:Evaluate the connection form on the fundamental vertical term] The connection form $\omega$ reproduces fundamental vector fields: for every $\xi\in\mathfrak g$, \begin{align*} \omega_r(\xi_P(r))=\xi \end{align*} at every $r\in P$. Applying this with $r=p=s_j(x)$ and $\xi=\eta_X$ gives \begin{align*} \omega_p((\eta_X)_P(p))=\eta_X. \end{align*} By the definition of $\eta_X$ and of the pullback $g^*\theta^L$, we have \begin{align*} \eta_X=\theta^L_{g(x)}(dg_x(X))=(g^*\theta^L)_x(X). \end{align*} [/step] [step:Combine the two evaluations to obtain the identity of local forms] Using the linearity of $\omega_p:T_pP\to\mathfrak g$ and the decomposition of $(ds_j)_x(X)$ from the first step, we obtain \begin{align*} (A_j)_x(X)=\omega_p((ds_j)_x(X)). \end{align*} Substituting the two evaluated terms gives \begin{align*} (A_j)_x(X)=\operatorname{Ad}(g(x)^{-1})(A_i)_x(X)+(g^*\theta^L)_x(X). \end{align*} Since $x\in U$ and $X\in T_xU$ were arbitrary, this proves the equality of $\mathfrak g$-valued $1$-forms \begin{align*} A_j=\operatorname{Ad}(g^{-1})A_i+g^*\theta^L \end{align*} on $U=U_{ij}$. Restoring the notation $g=g_{ij}$ gives \begin{align*} A_j=\operatorname{Ad}(g_{ij}^{-1})A_i+g_{ij}^*\theta^L. \end{align*} For a matrix Lie group, $\theta^L_g(dg)=g^{-1}dg$, so the same identity is written \begin{align*} A_j=\operatorname{Ad}(g_{ij}^{-1})A_i+g_{ij}^{-1}dg_{ij}. \end{align*} This is the asserted gauge transformation law for the local connection forms. [/step]