Show that the linear transformation with rank $m$ on $n$-dimensional subspace $V$ can be expressed as the sum of $m$ linear transformations with rank $1$.
Since any linear transformation can be represented as a matrix product, i.e $\mathbf x \mapsto A \mathbf x$, along with the property of rank of the matrices,
A rank-$k$ matrix can be written as the sum of $k$ rank-$1$ matrices.
This follows the conclusion.
The proof above is my attempt, and I think I might miss out something, or just have a proof for special case.
Any thought or suggestion would be appreciated.
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 $f_i(V) = \langle w_i\rangle$.