Is the parallel transport on an associated $G$-bundle given by $G$-action in a local $G$-bundle chart? Suppose we have a principal $G$-bundle $P \to M$ with $G$-atlas $(U_{\alpha}, \phi_{\alpha})_{\alpha \in A}$ and principal connection $\omega$, a left $G$-manifold $S$ (with not-necessarily effective action), and a smooth curve $c: I \to U_{\beta}$. Then we have a connection induced on the associated bundle $P \times_G S$. 
On $U_\beta$, the associated bundle looks like the trivial bundle $U_\beta \times S \to U_\beta$, and we denote the local principal connection form by $\omega_{\beta}: T U_\beta \to \mathfrak{g}$. Given a smooth curve $c: I \to U_\beta$ where $I=[0, 1]$, will the corresponding parallel transport be given by $G$-action? 
To be more precise, we can use the nonabelian fundamental theorem of calculus to integrate $c^* \omega_{\beta} : TI \to \mathfrak{g}$ to a curve $\widetilde{c}: I \to G$ with $\widetilde{c}(0) = e$ such that $\widetilde{c}^* \omega_G = c^* \omega_{\beta}$ where $\omega_G$ is the Maurer-Cartan form of $G$. In this case, is the parallel transport
$$\mathrm{Pt}(c, t): \{c(0)\}\times S \to \{c(t)\}\times S $$ 
in the bundle chart given by the formula $$\mathrm{Pt}(c,t)(-) = \widetilde{c}(t).(-)?$$
 A: The answer is yes. I'm not that familiar with the Maurer-Cartan formalism so I'll offer a perspective based directly on parallel transport.
Let $c \colon I \rightarrow M$ be a smooth curve with $c(0) = p$. By design, the parallel transport maps $\operatorname{Pt}^P_{c,0,t} \colon P_{c(0)} \rightarrow P_{c(t)}$ of $P$ are $G$-equivariant. How does the parallel transport on $E = P \times_G S$ is related to the parallel transport on $P$? Every element $\xi \in E_{p}$ is by definition an equivalence class $\xi = [\sigma_0, s_0]$ with $\sigma_0 \in P_p$ and $s_0 \in S$. The parallel transport on $E$ is then given by
$$ \operatorname{Pt}^E_{c,0,t}(\xi) = \operatorname{Pt}^E_{c,0,t}([\sigma_0, s_0]) = [\operatorname{Pt}^P_{c,0,t}(\sigma_0), s_0].$$
Namely, to parallel transport $\xi$ along $c$, we "represent it with respect to arbitrary frame" as $\xi = [\sigma, s_0]$, parallel transport the frame and keep the second component constant.
Now suppose you are given in advance a section $\sigma(t)$ of $P$ along $c$ which trivializes $P$ (along $c$). This section also gives us a trivialization of $E$ along $c$ by identifying $[\sigma(t), s]$ with $s$. How do the parallel transport maps look like with respect to this trivialization? We have
$$ s \cong [\sigma(0),s] \mapsto [\operatorname{Pt}^P_{c,0,t}(\sigma(0)), s] = [\sigma(t), g^{-1}(t) \cdot s] \cong g^{-1}(t)s $$
for a unique $g(t) \in G$ such that $\operatorname{Pt}^P_{c,0,t}(\sigma(0))g(t) = \sigma(t)$. Hence, the parallel transport maps with respect to such trivializations factor through a left $G$-action. 
Although I haven't checked the details, it seems reasonable that your $\tilde{c}(t)$ is precisely my $g^{-1}(t)$.
