This is how Axler defines a matrix
Suppose $T \in \mathcal{L}(V,W)$ and $v_1, \dots, v_n$ is a basis of
$V$, and $w_1, \dots, w_n$ is a basis of $W$. The matrix of $T$
with respect to these bases is the $m$-by-$n$ matrix $\mathcal{M}(T)$
whose entries $A_{j,k}$ are defined by$$T(v_k) = A_{1,k} w_1 + \dots + A_{m,k}w_m$$
I'm not sure what this means. He gives an example:
$$T(x,y) = (x+3y, 2x+5y, 7x+9y)$$.
With respect to the standard bases, the matrix is
$$\begin{pmatrix}
1 & 3 \\
2 & 5 \\
7 & 9
\end{pmatrix}$$
That's obvious through my conventional understanding of matrices. Suppose my bases are $(2,0),(0,8)$ for $\mathbb{R}^2$ and the standard basis for $\mathbb{R}^3$. I have $T(2,0) = (2,4,14)$ and $T(0,8)=(24,40,72)$. What would my matrix be then? Would it be the following?
$$\begin{pmatrix}
2 & 24 \\
4 & 40 \\
14 & 72
\end{pmatrix}$$
However, with this matrix, how would I carry out matrix multiplications as I usually do? Because
$$\begin{pmatrix}
2 & 24 \\
4 & 40 \\
14 & 72
\end{pmatrix} \begin{pmatrix}
1 \\
2
\end{pmatrix} \neq T((1,2))$$
Best Answer
In the next section of the book, we have the following result: $$\mathcal{M}(Tv) = \mathcal{M}(T) \mathcal{M}(v).$$
As you have calculated, $T(2,0) = (2,4,14)$ and $T(0,8) = (24,40,72)$. Since we are looking at the standard basis in $\mathbb{R}^3$, the matrix of $T$ is, as you showed, $$\mathcal{M}(T) = \begin{pmatrix}2 & 24\\ 4 & 40\\ 14 & 72\end{pmatrix}.$$
The matrix of the vector $v = (1,2)$ with respect to the basis $\{ (2,0), (0,8) \}$ is $$\mathcal{M}(v) = \begin{pmatrix}1/2 \\ 1/4 \end{pmatrix}.$$
And, since $T(1,2) = (7,12,25)$, the matrix of $T(1,2)$ with respect to the standard basis is $$\mathcal{M}(Tv) = \begin{pmatrix}7\\ 12\\ 25\end{pmatrix}.$$
As you can verify, $\mathcal{M}(Tv) = \mathcal{M}(T) \mathcal{M}(v).$
Perhaps your confusion arises because we often work with the standard bases. For example, if we are working with the standard basis in $\mathbb{R}^2$ and also the standard basis in $\mathbb{R}^3$, then $\mathcal{M}(v) = v$ and $\mathcal{M}(Tv) = Tv$. Hence, we have $Tv = \mathcal{M}(T)v$. So indeed we may "just multiply $v$ by the matrix" to get $Tv$.