I understand what it means for vectors, i.e.
$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & -3 & -1 \\ -1 & 2 & -2 \end{pmatrix} \begin{pmatrix} 4/3 \\ 1/6 \\ -1/2 \end{pmatrix}, $$
but when it comes to representations of linear maps, i.e.
$$\begin{pmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},$$
my intuition fails me. I have two problems with this:
Representations depend on 2 bases, the one from the original space and the one from the target space, so which one changes?
I suppose that it is very similar to the vector case, but I don't know how to proceed with the adaptation. A disregarded "thanks".
Let $L:U \rightarrow V $ be a linear map. Then, given basis of $\{u_i\}$ of $U$, $\{v_i\}$ of $V$, the matrix $M_L$ which represents $L$ in the given bases has components $(M_L)^i_{\ j}= v^i (L(u_j))$, where $\{v^i\}$ is the dual basis of $\{v_i\}$. There is a reason for placing one index above and one below in $M_L$ but ignore it and just think of $(M_L)_{ij}$ if you are not familiar with it - or ask if you are curious.
Perhaps another wording will be clearer: the matrix associated to a linear map $L:U\rightarrow V$ with respect to the bases above has as $j$th column the components of the image of the $j$th basis vector of the domain vector space $U$ $L(u_j)$ with respect to the basis $\{v_i\}$ of the target vector space $V$.