The tangent space of $\mathrm{Aut}(T_eG)$

If $G$ is a manifold of dimension $n$, then $T_eG$ is just a $n$-dimensional real vector space, so that after choosing a basis of $T_eG$, we can make an identification $T_eG\cong \mathbb{R}^n$. The endomorphisms of $\mathbb{R}^n$ (i.e., the linear maps from $\mathbb{R}^n$ to itself) can be identified with the $n\times n$ matrices, and hence so can the endomorphisms of $T_eG$. The automorphisms of $\mathbb{R}^n$ are the invertible, a.k.a. "non-singular" matrices. The space of $n\times n$ matrices has its own topology, and in this topology, the invertible matrices form an open set; that is covered in this older answer of mine.

As to your second question, for any real vector space $V$ of finite dimension $n$, and any $v\in V$, there is a natural identification of $T_vV$ with $V$ (this is Prop 3.8 in Lee's Introduction to Smooth Manifolds, 1st ed). That gives us an identification $$T_\mathrm{id}(\mathrm{End}(T_eG))\cong \mathrm{End}(T_eG)$$ where $\mathrm{id}\in\mathrm{End}(T_eG)$ is the identity map from $T_eG$ to itself (I assume this is what you intended in your problem statement). Additionally, for any manifold $M$, any open set $U\subseteq M$, and any $p\in U$, there is a natural identification of $T_pU$ with $T_pM$ (this is Prop 3.7, ibid.). Thus, we have a sequence of identifications $$T_\mathrm{id}(\mathrm{Aut}(T_eG))\cong T_\mathrm{id}(\mathrm{End}(T_eG))\cong \mathrm{End}(T_eG).$$