I apologize that my question is a bit long, but when I try to repeat the first process in a different problem, I am unable to and am wondering if this is tied to diagonalizability.
Let $T: \mathbb{R} \to \mathbb{R}$ be the linear transformation with respect to the standard basis represented by the matrix $$A=\begin{pmatrix}
3 & 1 \\ 2 &2
\end{pmatrix}.$$
Let's use change-of-basis, to find a new matrix $A'$ to represent this linear transformation with respect to $$\vec{v}_1=\begin{pmatrix}
1 \\ 1
\end{pmatrix} \quad \text{and} \quad \vec{v}_2=\begin{pmatrix}
-1 \\ 2
\end{pmatrix}$$
The matrix that represents $T$ with respect to the ordered basis $\mathcal{B}'=\{\vec{v_1},\vec{v_2}\}$ is the diagonal matrix $$A'=\begin{pmatrix}
4 & 0 \\ 0 & 1
\end{pmatrix}.$$
We can see that if write the first component of the old basis $\vec{e_1}$ in terms of the new basis vectors, and multiply them by $A'$, you get the first column of $A$.
$$\vec{e}_1=\frac{2}{3}\vec{v}_1-\frac{1}{3}\vec{v}_2$$
$$T(\vec{e}_1)=T(\frac{2}{3}\vec{v}_1-\frac{1}{3}\vec{v}_2)=\frac{2}{3}T(\vec{v}_1)-\frac{1}{3}T(\vec{v}_2)=\begin{pmatrix}
3 \\ 2
\end{pmatrix}$$
I try then to repeat this same process with a different problem where the new standard matrix is $$A=\begin{pmatrix}
1 & 5 \\ 2 & -2
\end{pmatrix}.$$
The new basis is $$\vec{v}_1=\begin{pmatrix}
2 \\ 3
\end{pmatrix} \quad \text{and} \quad \vec{v}_2=\begin{pmatrix}
1 \\ 2
\end{pmatrix}$$ $$A'=\begin{pmatrix}
36 & 24 \\ -55 & -37
\end{pmatrix}$$
When I try to find $T(\vec{e}_1)$ as in the previous example, I do not find that the solution is the first column of $A$. Does this process not work in the second problem because $A$ is not a diagonal matrix?
The reason why the first problem works is because $\vec v_1$ and $\vec v_2$ are eigenvectors of $A$: in particular, $A\vec v_1 = 4\vec v_1$ and $A\vec v_2 = 1\cdot \vec v_2$.
In your second example, the $\vec v$'s are not eigenvectors of $A$. (However, $\displaystyle{-1\choose 1}$ and $\displaystyle{5\choose 2}$ are.)