Note: Please do not solve the exercise for me, I would very much like to do it myself.
The following is exercise 27.5 from Loring Tu's "Differential Geometry".
Let $\phi_\alpha:\pi^{-1}U_\alpha\to U_\alpha\times G$ given by $\phi_\alpha(p)=(\pi(p),g_\alpha(p))$ be a trivialisation of $\pi^{-1}U_{g\color{red}{a}}$ in a principal bundle $P$. Let $A\in\mathfrak{g}$ and $\bar{A}$ the fundamental vector field on $P$ that it induces. Prove that $dg_\alpha(\bar{A}_p)=dl_{g_\alpha(p)}(A)$.
I have several questions about this exercise. Firstly, the $a$ in red is presumably a typo, because $a$ isn't mentioned anywhere else. I am guessing this should be $\alpha$, but then what is the $g$ doing there? My second question pertains to the differential itself. It isn't clear to me from where to where this mapping goes. I think the target space is $T_{g_\alpha(p)}G$, but then where does it map from?
This community wiki solution is intended to clear the question from the unanswered queue.
Question 1: It is $\pi^{-1}U_\alpha$. See Ted Shifrin's comment.
Question 2: You have $\phi_\alpha:\pi^{-1}U_\alpha\to U_\alpha\times G$ given by $\phi_\alpha(p)=(\pi(p),g_\alpha(p))$. Thus $g_\alpha : \pi^{-1}U_\alpha\to G$ and $d_pg_\alpha : T_p \pi^{-1}U_\alpha \to T_{g_\alpha(p)}G$. See A. Bellmunt's comment.