[proofplan]
Represent the section $s$ by the corresponding $\rho$-equivariant function $f_s:P\to V$, so that $f_s(\sigma_i(x))=s_i(x)$ on the chosen trivialising [open set](/page/Open%20Set). The induced covariant derivative differentiates this equivariant function only in horizontal directions. We decompose $(d\sigma_i)_x(X)$ into its horizontal part plus the vertical fundamental vector generated by $A_i(x)(X)$, compute the vertical derivative using the convention $f_s(pg)=\rho(g^{-1})f_s(p)$, and then rewrite the resulting expression in terms of $s_i$ and $A_i$.
[/proofplan]
[step:Pass from the section to its equivariant representative]
Let
\begin{align*}
f_s:P\to V
\end{align*}
be the smooth equivariant function corresponding to $s$, defined by the condition that
\begin{align*}
s(\pi(p))=[p,f_s(p)]
\end{align*}
for every $p\in P$. Because the associated bundle uses the [equivalence relation](/page/Equivalence%20Relation) $(p,v)\sim(pg,\rho(g^{-1})v)$, the corresponding equivariance condition is
\begin{align*}
f_s(pg)=\rho(g^{-1})f_s(p)
\end{align*}
for every $p\in P$ and $g\in G$.
For $x\in U_i$, the defining relation for the local representative $s_i:U_i\to V$ gives
\begin{align*}
s(x)=[\sigma_i(x),s_i(x)].
\end{align*}
Comparing this with $s(x)=[\sigma_i(x),f_s(\sigma_i(x))]$ yields
\begin{align*}
f_s(\sigma_i(x))=s_i(x).
\end{align*}
Thus the composite map
\begin{align*}
f_s\circ\sigma_i:U_i\to V
\end{align*}
is exactly $s_i$.
[guided]
The section $s$ of the associated bundle can be encoded by a function on the principal bundle. Define
\begin{align*}
f_s:P\to V
\end{align*}
by requiring
\begin{align*}
s(\pi(p))=[p,f_s(p)]
\end{align*}
for each $p\in P$. This definition is compatible with changing $p$ along the right $G$-orbit only if $f_s$ satisfies the equivariance law
\begin{align*}
f_s(pg)=\rho(g^{-1})f_s(p).
\end{align*}
The inverse on $g$ is forced by the convention used in the associated bundle: $(p,v)$ and $(pg,\rho(g^{-1})v)$ represent the same vector.
Now restrict this relation to the local section $\sigma_i:U_i\to P$. The local representative $s_i:U_i\to V$ is defined by
\begin{align*}
s(x)=[\sigma_i(x),s_i(x)].
\end{align*}
On the other hand, the equivariant function gives
\begin{align*}
s(x)=[\sigma_i(x),f_s(\sigma_i(x))].
\end{align*}
Both expressions use the same point $\sigma_i(x)$ in the principal fibre, so the vector component is the same:
\begin{align*}
f_s(\sigma_i(x))=s_i(x).
\end{align*}
Therefore
\begin{align*}
f_s\circ\sigma_i=s_i.
\end{align*}
This identity is the bridge between the global principal-bundle definition of the induced connection and the local formula in the trivialisation over $U_i$.
[/guided]
[/step]
[step:Decompose the derivative of the local section into horizontal and vertical parts]
Fix $x\in U_i$ and $X\in T_xM$. Define
\begin{align*}
p:=\sigma_i(x)
\end{align*}
and
\begin{align*}
\xi:=A_i(x)(X)\in\mathfrak g.
\end{align*}
Let $\xi_P(p)\in T_pP$ denote the fundamental vertical vector generated by $\xi$, namely
\begin{align*}
\xi_P(p)=\frac{d}{dt}\bigg|_{t=0}p\exp(t\xi).
\end{align*}
Since $A_i=\sigma_i^*\omega$, we have
\begin{align*}
\xi=A_i(x)(X)=\omega_p((d\sigma_i)_x(X)).
\end{align*}
The connection form satisfies $\omega_p(\xi_P(p))=\xi$, so
\begin{align*}
\omega_p((d\sigma_i)_x(X)-\xi_P(p))=0.
\end{align*}
Thus
\begin{align*}
H:= (d\sigma_i)_x(X)-\xi_P(p)\in T_pP
\end{align*}
is horizontal. Also
\begin{align*}
d\pi_p(H)=d\pi_p((d\sigma_i)_x(X))=X,
\end{align*}
because $\pi\circ\sigma_i=\operatorname{id}_{U_i}$ and $d\pi_p(\xi_P(p))=0$.
[/step]
[step:Compute the horizontal derivative defining the induced connection]
By definition of the induced covariant derivative, the local representative of $\nabla_X^E s$ at $x$ is obtained by differentiating $f_s$ in the horizontal lift of $X$ through $p=\sigma_i(x)$:
\begin{align*}
(\nabla_X^E s)_i(x)=d(f_s)_p(H).
\end{align*}
Using the horizontal decomposition from the previous step,
\begin{align*}
d(f_s)_p(H)=d(f_s)_p((d\sigma_i)_x(X))-d(f_s)_p(\xi_P(p)).
\end{align*}
The first term is computed by the chain rule applied to $f_s\circ\sigma_i=s_i$:
\begin{align*}
d(f_s)_p((d\sigma_i)_x(X))=d(s_i)_x(X).
\end{align*}
For the vertical term, use the curve
\begin{align*}
\gamma:(-\varepsilon,\varepsilon)\to P,\qquad \gamma(t)=p\exp(t\xi)
\end{align*}
for sufficiently small $\varepsilon>0$. Then $\gamma'(0)=\xi_P(p)$, and equivariance gives
\begin{align*}
f_s(\gamma(t))=\rho(\exp(-t\xi))f_s(p).
\end{align*}
Differentiating at $t=0$ gives
\begin{align*}
d(f_s)_p(\xi_P(p))=-\rho_*(\xi)f_s(p).
\end{align*}
Therefore
\begin{align*}
d(f_s)_p(H)=d(s_i)_x(X)+\rho_*(\xi)f_s(p).
\end{align*}
Since $\xi=A_i(x)(X)$ and $f_s(p)=s_i(x)$, this becomes
\begin{align*}
(\nabla_X^E s)_i(x)=d(s_i)_x(X)+\rho_*(A_i(x)(X))s_i(x).
\end{align*}
[guided]
The induced connection differentiates the equivariant representative $f_s:P\to V$ only in horizontal directions. For the fixed point $x\in U_i$ and tangent vector $X\in T_xM$, set
\begin{align*}
p:=\sigma_i(x)
\end{align*}
and
\begin{align*}
\xi:=A_i(x)(X)\in\mathfrak g.
\end{align*}
From the previous step, the horizontal lift of $X$ through $p$ is
\begin{align*}
H=(d\sigma_i)_x(X)-\xi_P(p),
\end{align*}
where
\begin{align*}
\xi_P(p)=\frac{d}{dt}\bigg|_{t=0}p\exp(t\xi)
\end{align*}
is the fundamental vertical vector associated to $\xi$.
By definition of $\nabla^E$, the representative of $\nabla_X^E s$ in the frame determined by $\sigma_i$ is
\begin{align*}
(\nabla_X^E s)_i(x)=d(f_s)_p(H).
\end{align*}
Substitute the decomposition of $H$:
\begin{align*}
d(f_s)_p(H)=d(f_s)_p((d\sigma_i)_x(X))-d(f_s)_p(\xi_P(p)).
\end{align*}
The first term is ordinary differentiation of the local representative. Indeed, the identity $f_s\circ\sigma_i=s_i$ implies, by the chain rule,
\begin{align*}
d(f_s)_p((d\sigma_i)_x(X))=d(s_i)_x(X).
\end{align*}
The second term is where the sign enters. To compute it, consider the smooth curve
\begin{align*}
\gamma:(-\varepsilon,\varepsilon)\to P,\qquad \gamma(t)=p\exp(t\xi).
\end{align*}
Its derivative at $0$ is exactly $\xi_P(p)$. Since $f_s$ is equivariant with the convention $f_s(pg)=\rho(g^{-1})f_s(p)$, we get
\begin{align*}
f_s(\gamma(t))=f_s(p\exp(t\xi))=\rho(\exp(-t\xi))f_s(p).
\end{align*}
Differentiating at $t=0$ gives
\begin{align*}
d(f_s)_p(\xi_P(p))=-\rho_*(\xi)f_s(p).
\end{align*}
The minus sign comes from differentiating $\exp(-t\xi)$ rather than $\exp(t\xi)$.
Now insert this into the horizontal derivative:
\begin{align*}
d(f_s)_p(H)=d(s_i)_x(X)-(-\rho_*(\xi)f_s(p)).
\end{align*}
Hence
\begin{align*}
d(f_s)_p(H)=d(s_i)_x(X)+\rho_*(\xi)f_s(p).
\end{align*}
Finally, $\xi=A_i(x)(X)$ by definition of the local connection form, and $f_s(p)=f_s(\sigma_i(x))=s_i(x)$ by the local representative identity. Therefore
\begin{align*}
(\nabla_X^E s)_i(x)=d(s_i)_x(X)+\rho_*(A_i(x)(X))s_i(x).
\end{align*}
[/guided]
[/step]
[step:Conclude the local one form identity]
The preceding identity holds for every $x\in U_i$ and every $X\in T_xM$. Therefore the $V$-valued one-form on $U_i$ representing $\nabla^E s$ is
\begin{align*}
(\nabla^E s)_i=ds_i+\rho_*(A_i)s_i.
\end{align*}
This is the asserted local formula for the covariant derivative induced by the principal connection $\omega$.
[/step]
