I am trying to work out how one obtains the adjoint representation of a Lie algebra from the adjoint representation of the Lie group. I apologise if this is a trivial question but I am a physicist so I am rusty.
Consider a Lie group $G$ and its corresponding Lie algebra $\mathfrak{g}$. The adjoint representation of $G$ is given by
$$ \mathrm{Ad}: G \rightarrow \mathrm{Aut}(\mathfrak{g}).$$
According to the Wiki article, I can obtain the adjoint representation $\mathrm{ad}$ of the Lie algebra from the pushforward of $\mathrm{Ad}$:
$$ \mathrm{ad} : = \mathrm{Ad}_* : \mathfrak{g} \rightarrow \mathrm{Der}(\mathfrak{g}), $$
where $\mathrm{Der}(g)$ is the derivation algebra.
Now, from my knowledge of the pushforward, I would say that the pushforward maps vectors in one tangent space, to vectors in another. So I would expect $\mathrm{Ad}_*$ to act as
$$ \mathrm{Ad}_* : \mathfrak{g} \rightarrow T_I(\mathrm{Aut}(\mathfrak{g})), $$
where $I$ is the identity of $\mathrm{Aut}(\mathfrak{g})$. What I do not understand is why the elements of $T_I(\mathrm{Aut}(\mathfrak{g}))$ could be interpreted as operators on $\mathfrak{g}$. In other words, I do not understand why $T_I(\mathrm{Aut}(\mathfrak{g})) \cong \mathrm{Der}(\mathfrak{g})$. From a differential geometry perspective, tangent vectors are objects which act on functions $f : M \rightarrow \mathbb{R}$, so where do we deduce that these tangent vectors of $T_I(\mathrm{Aut}(\mathfrak{g}))$ are in fact operators on $\mathfrak{g}$?
Theorem 2.3 of this states that $ A \in T_I(\mathrm{Aut}(\mathfrak{g})) $ then for $Y \in \mathfrak{g}$ it acts as
$$ A(Y) = \frac{d}{dt} \mathrm{Ad}(\exp tX) Y \bigg|_{t=0}$$
where $A = \mathrm{ad}(X)$. I do not understand whether this is a definition or can be deduced purely from $\mathrm{ad}$, but I do not know why this is how $\mathrm{ad}$ should act. It also seems like an abuse of notation here too, should this not be acting on some test function $f$?