I think your difficulty is purely "conceptual". A vector space is ANY set that obeys the axioms of a vector space (such a definition assumes an associated "field of scalars"). In particular, over a given field $F$, the set:
$\text{Hom}_F(U,V)$ = all linear transformations $U \to V$ is a vector space.
So "matrices" are vectors, too! Look, we can add them: if $A = (a_{ij}), B = (b_{ij})$ then $A + B = (c_{ij})$, where for each $i,j: c_{ij} = a_{ij} + b_{ij}$, and we can "multiply by a scalar":
$rA = (ra_{ij})$ (sometimes this is written as $(rI)A$).
One way to "ease the conceptual transition" is called the "vectorization" of matrices: we just string the columns "head-to-toe" into one long column, so that:
$A = \begin{bmatrix}1&2\\0&3\end{bmatrix}$ becomes:
$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix}$, transforming an element of $\text{Mat}_2(F)$ into an element of $F^4$ (you can take $F = \Bbb R$, for concreteness, if you primarily deal with real vector spaces).
Seen this way, it becomes clear that the second coordinate of your basis vectors is "unnecessary baggage", as it is always $0$. This is no different than identifying the subspace:
$U = \{(x,0,y,z) \in \Bbb R^4\}$ with $\Bbb R^3 = \{(x,y,z):x,y,z \in \Bbb R\}$
It should be clear that $\phi:U \to \Bbb R^3$ given by $\phi(x,0,y,z) = (x,y,z)$ is a bijective linear transformation.
So, in your situation, you want to find $a,b,c$ such that:
$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix} = a\begin{bmatrix}1\\0\\0\\0\end{bmatrix} + b\begin{bmatrix}0\\0\\1\\0\end{bmatrix} + c\begin{bmatrix}0\\0\\1\\1\end{bmatrix}$
Which is just a system of 3 linear equations in 3 unknowns:
$a = 1\\b+c = 2\\c = 3$
that you should be proficient in solving by now.
Best Answer
The matrix of a transformation is the matrix that turns the vector of coordinates of the input into the vector of coordinates of the output in certain bases. Coordinate vectors are always column vectors of some $\mathbb{R}^n$.
For example: In the standard basis $\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\}$, of the space of matrices. The coordinate vector of the matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, is the column $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$.
In your case the transformation is $$T(\begin{bmatrix}a&b\\c&d\end{bmatrix})=\begin{bmatrix}2ia&b+ci\\c+bi&2id\end{bmatrix}$$
When you compute the matrix of this transformation in the standard basis you get $$A_T:=\begin{bmatrix}2i&0&0&0\\0&1&i&0\\0&i&1&0\\0&0&0&2i\end{bmatrix}$$
What this means is that when you multiply the vector of coordinates $$\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$ of some vector $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ of your space, by $A_T$ you get the vector of coordinates $$\begin{bmatrix}2ia\\b+ci\\c+bi\\2di\end{bmatrix}=A_t\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$
of the output of $T$. This is, the vector of coordinates of the vector $\begin{bmatrix}2ai&b+ci\\c+bi&2di\end{bmatrix}$ in the standard basis.