I have been studying for an exam and I managed to forget elementary stuff concerning pushforwards and I want to clarify some intuition. Let's say I have a Lie group $G$ and its Lie algebra $\mathfrak{g}$. I will denote conjugation with $g \in G$ as $I_g$, so $I_g h = ghg^{-1}$ for some $h \in G$. If I have the following
$$Ad_g(X) = I_{g*}\frac{d}{dt}\bigg|_{t=0} \exp(tX) =\frac{d}{dt}\bigg|_{t=0}g \exp(tX)g^{-1}, $$
where $X \in \mathfrak{g}$. I understand the first equality, which comes simply from the fact how the exponential and the adjoint representation are defined, however I can't grasp why the second equality works. If I look at the definition of the pushforward which would be for some $\phi: M \to N$, where $M, N$ are smooth manifolds and $Y \in T_pM$ for $p \in M$:
$$\phi_*(Y)f = Y(f \circ \phi).$$
How do these two things come together? I feel like I should be thinking of $Y$ as "derivatives of curves that pass trough $p$" and somehow rewriting this condition for pushforward. Does
$$\phi_*(Y) =\frac{d}{dt}\bigg|_{t=0}\phi(\gamma(t))$$
hold if $\gamma$ is integral curve of $Y$, such that $\gamma(p)$ = 0. If it does hold, why is that the case? It looks like I am missing something very trivial in all of this.