As I see it, the author says that $[Tv]_{e} = A[v]_{e}$ in the last paragraph. How do I see that ?
I think I've jusitied the first entry in $[Tv]_{e} = A[v]_{e}$ as follows
\begin{align*} \langle \text{T}v,e_{1}\rangle &= \langle \text{T}e_{1},e_{1}\rangle\langle v,e_{1}\rangle + \cdots + \langle \text{T}e_{n},e_{1}\rangle\langle v,e_{n}\rangle\\ &=\langle\langle \text{T}e_{1},e_{1}\rangle v,e_{1}\rangle\rangle\\ &=\langle \text{T}v,e_{1}\rangle. \end{align*}
Is there any other way to see it, or interpret it ?
The idea is this - there is a correspondence between matrices in $\mathbb{F}^{n\times n}$ and linear operators in $\mathcal{L}(V,V)$.
To every linear operator $T\in \mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $\mathcal{B} = \{e_i\}$ of $V$, and consider the matrix $T_{\mathcal{B}}$ whose columns are the vectors $\{T(e_i)\}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $\mathcal{B}'$, then how are the two matrices $T_{\mathcal{B}}$ and $T_{\mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that $$ T_{\mathcal{B}'} = PT_{\mathcal{B}}P^{-1} $$