Let the transformation $T$ be from $R^n \to R^m$. We will need bases for each of these spaces, let them be $B_n = \{e_{1n}, e_{2n}... e_{nn}\}$ and $B_m = \{e_{1m}, e_{2m}... e_{mm}\}$ respectively.
Now, any vector $v$ can be expressed as the following
$$v = \sum_1^na_ie_{in}$$
$$\implies T(v) = \sum_i^na_iT(e_{in})$$
To complete the matrix representation, we need to express each $T(e_{in})$ in the basis of the $m$-space
Hence, let $T(e_{in}) = \sum_{k=1}^mb_{ik}e_{km}$
Therefore
$$\implies T(v) = \sum_{i=1}^na_i\sum_{k=1}^mb_{ik}e_{km}$$
Now, we consider the matrix representation of $T$, we express $v$ as a column vector in $R^{n \times 1}$
$$v = \begin{bmatrix}a_1 \\ a_2 \\ . \\. \\. \\a_n\end{bmatrix}$$
Hence, $T(v)$ can be thought of as the sum of $m$ vectors in $R^{m \times 1}$, weighted by the $v$ column scalars. Therefore, we pre-multiply by the column wise representation of $T(e_{in})$ in the basis $B_m$, which is given by scalars $b_{ik}$ as defined above
$$[T] = \begin{bmatrix} b_{11} & b_{21} & b_{31} & ... & b_{n1} \\ b_{12} & b_{22} & b_{32} &...& b_{n2} \\ . & .& . \\ .&.&.&.\\b_{1m} & b_{2m} & b_{3m} &...&b_{nm} \end{bmatrix}$$
Best Answer
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $f\colon V\to W$ be a linear map such that $\dim(f(V))=m$. Pick a basis $w_1,\dots,w_m$ of $f(V)$ and define projections \begin{align*} \pi_j \colon\quad\quad f(V)&\longrightarrow f(V), \\ \sum_i \lambda_i w_i &\longmapsto \lambda_j w_j \end{align*} for $j=1,\dots,m$. Note that $\pi_1+\cdots+\pi_m = \operatorname{id}_{f(V)}$.
Hence, defining $f_i\colon V\to W$ by $f_i(v)=\pi_i(f(v))$ we have $$ f = f_1+\cdots + f_n. $$ Each $f_i$ is of rank $1$ since $f_i(V) = \langle w_i\rangle$.