[proofplan]
Write $v$ in the ordered basis $B$, and use linearity of $T$ to express $T(v)$ as the same linear combination of the vectors $T(b_j)$. Then expand each $T(b_j)$ in the ordered basis $C$; the resulting coordinate vectors are exactly the columns of the matrix $[T]_{C \leftarrow B}$. Matrix-vector multiplication forms the same linear combination of these columns, so its entries are precisely the coordinates of $T(v)$ in the basis $C$.
[/proofplan]
[step:Expand $v$ in the ordered basis $B$]
Let $v \in V$ be arbitrary. Since $B=(b_1,\ldots,b_n)$ is a basis of $V$, there exist unique scalars $a_1,\ldots,a_n \in k$ such that
\begin{align*}
v = \sum_{j=1}^{n} a_j b_j.
\end{align*}
By definition of the coordinate column vector with respect to $B$,
\begin{align*}
[v]_B =
\begin{pmatrix}
a_1
\end{align*}
\begin{align*}
\vdots
\end{align*}
\begin{align*}
a_n
\end{pmatrix}.
\end{align*}
[/step]
[step:Apply linearity of $T$ to the coordinate expansion]
Since $T: V \to W$ is $k$-linear, applying $T$ to the expansion of $v$ gives
\begin{align*}
T(v) = \sum_{j=1}^{n} a_j T(b_j).
\end{align*}
[guided]
We now transfer the coordinate expansion from $V$ to $W$. The point of using a basis of the domain is that every vector $v \in V$ has a unique expression
\begin{align*}
v = \sum_{j=1}^{n} a_j b_j.
\end{align*}
The scalars $a_1,\ldots,a_n \in k$ are exactly the entries of the coordinate vector $[v]_B$.
Because $T$ is $k$-linear, it preserves finite sums and scalar multiplication. Therefore
\begin{align*}
T(v) = T\left(\sum_{j=1}^{n} a_j b_j\right).
\end{align*}
Using additivity and homogeneity of $T$, this becomes
\begin{align*}
T(v) = \sum_{j=1}^{n} a_j T(b_j).
\end{align*}
This identity is the key bridge between the abstract map $T$ and its matrix: the same coefficients that describe $v$ in the domain basis $B$ describe $T(v)$ as a linear combination of the images of the basis vectors.
[/guided]
[/step]
[step:Write the images of the basis vectors in the ordered basis $C$]
For each $j \in \{1,\ldots,n\}$, since $C=(c_1,\ldots,c_m)$ is a basis of $W$, there exist unique scalars $\alpha_{1j},\ldots,\alpha_{mj} \in k$ such that
\begin{align*}
T(b_j) = \sum_{i=1}^{m} \alpha_{ij} c_i.
\end{align*}
By definition of the matrix of $T$ with respect to $B$ and $C$, the matrix $[T]_{C \leftarrow B}$ is
\begin{align*}
[T]_{C \leftarrow B} = (\alpha_{ij})_{1 \leq i \leq m,\ 1 \leq j \leq n}.
\end{align*}
Equivalently, its $j$th column is
\begin{align*}
[T(b_j)]_C =
\begin{pmatrix}
\alpha_{1j}
\end{align*}
\begin{align*}
\vdots
\end{align*}
\begin{align*}
\alpha_{mj}
\end{pmatrix}.
\end{align*}
[/step]
[step:Compare matrix multiplication with the coordinate expansion of $T(v)$]
Substituting the $C$-basis expansion of each $T(b_j)$ into the expression for $T(v)$ gives
\begin{align*}
T(v) = \sum_{j=1}^{n} a_j \sum_{i=1}^{m} \alpha_{ij} c_i.
\end{align*}
Rearranging the finite sums in the vector space $W$ gives
\begin{align*}
T(v) = \sum_{i=1}^{m} \left(\sum_{j=1}^{n} \alpha_{ij} a_j\right)c_i.
\end{align*}
By uniqueness of coordinates in the basis $C$,
\begin{align*}
[T(v)]_C =
\begin{pmatrix}
\sum_{j=1}^{n} \alpha_{1j}a_j
\end{align*}
\begin{align*}
\vdots
\end{align*}
\begin{align*}
\sum_{j=1}^{n} \alpha_{mj}a_j
\end{pmatrix}.
\end{align*}
On the other hand, by the definition of matrix-vector multiplication over $k$,
\begin{align*}
[T]_{C \leftarrow B}[v]_B =
(\alpha_{ij})_{1 \leq i \leq m,\ 1 \leq j \leq n}
\begin{pmatrix}
a_1
\end{align*}
\begin{align*}
\vdots
\end{align*}
\begin{align*}
a_n
\end{pmatrix}
=
\begin{pmatrix}
\sum_{j=1}^{n} \alpha_{1j}a_j
\end{align*}
\begin{align*}
\vdots
\end{align*}
\begin{align*}
\sum_{j=1}^{n} \alpha_{mj}a_j
\end{pmatrix}.
\end{align*}
Therefore
\begin{align*}
[T(v)]_C = [T]_{C \leftarrow B}[v]_B.
\end{align*}
Since $v \in V$ was arbitrary, the identity holds for every $v \in V$.
[/step]
