Let $v_1, \ldots, v_n$ be $n$ linearly independent vectors in $\mathbb{R}^n$, the real Euclidean $n$-space, let $e_1, \ldots, e_n$ be the unit coordinate vectors in $\mathbb{R}^n$; that is, let $$e_1 \colon= (1, 0, \ldots, 0),$$ $$e_2 \colon= (0, 1, \ldots, 0),$$ and so on $$e_n \colon= (0, 0, \ldots, 1).$$
Then there is a unique linear transformation $T \colon \mathbb{R}^n \to \mathbb{R}^n$ such that $T(v_k) = e_k$ for each $k = 1, \ldots, n$.
How can we find an $n \times n$ matrix $B$ such that $v_k B = e_k$, where $v_k$ and $e_k$ are to be regarded as a row matrices for each $k = 1, \ldots n$?
Theorem 2.12 in Apostol says, if $e_1,\dots,e_n$ is any basis (not just the standard basis) for a vector space, and $u_1,\dots,u_n$ are any vectors, there is a unique linear transformation $T$ such that $Te_k = u_k$.
Apply this statement to the basis $v_k$ (by Thm 1.7, $v_k$ form a basis): there exists a unique linear transformation $T$ such that $$ Tv_k = e_k, $$ and $B$ is just the matrix associated with this linear transformation in this basis (such a matrix always exists, see section 2.10).
The critical part is that the $v_k$ form a basis; if they didn't, there wouldn't be a $T$ (and so no associated $B$) such that $Tv_k = e_k$, and the whole argument doesn't get off the ground.