Let $A = \sum_{i=1}^k \vec{u}_i \vec{v}_i^T$, where $\vec{u}_i \in \mathbb{R}^m$ and $\vec{v}_i \in \mathbb{R}^n$. Assume that $\{ \vec{u}_i \}_{i=1}^k$ are linearly independent and $\{ \vec{v}_i \}_{i=1}^k$ are linearly independent. Assume $m > n > k$.
What is the rank of $A$?
Let $A_i = \vec{u}_i \vec{v}_i^T$ so that $A = \sum_{i=1}^k A_i$.
Column $j$ of $A_i$ is $(\vec{v}_i)_j \vec{u}_i$. The rank of each $A_i$ is one since all columns are multiples of a single vector $\vec{u}_i$.
Column $j$ of $A$ is $\sum_{i=1}^k (\vec{v}_i)_j \vec{u}_i$. It seems the columns of $A$ may or may not be linearly independent. Is that correct? Then the rank can be from $1$ to $k$. The wording of the question suggests that the rank of $A$ is $k$ but I don't see why this is necessarily the case.
On the one hand, for any $v \in \mathbf{R}^m$, we have that $$ Av = \sum_{i=1}^k (v_i^\mathrm{T}v) u_i \in \mathrm{span}\{u_1, \dotsc, u_k\}, $$ so that $\mathrm{Im}(A) \subseteq \mathrm{span}\{u_1, \dotsc, u_k\}$
On the other hand, let $u = \sum_{i=1}^k a_i u_i$ for $a_i \in \mathbf{R}$. Let $V$ be the $k\times n$ matrix whose $i$th row is $v_i^{\mathrm{T}}$, Since the $v_i$ are linearly independent, this matrix has rank $k$ and so is surjective. In particular, there is a $w\in \mathbf{R}^m$ that solves $Vw = a$, where $a$ is the vector whose $i$th entry is $a_i$. Equating entries, this means that $(Vw)_i = v_i^{\mathrm{T}} w = a_i$
But then, we have that $$ Aw = \sum_{i=1}^k (v_i^\mathrm{T}w) u_i = \sum_{i=1}^k a_i u = u, $$ so that $\mathrm{span}\{u_1, \dotsc, u_k\} \subseteq \mathrm{Im}(A)$.
Hence, the rank of $A$ is $$ \mathrm{\dim}\, \mathrm{Im}(A) = \mathrm{\dim}\, \mathrm{span}\{u_1, \dotsc, u_k\}, $$ which is $k$ since th $u_i$ are linearly independent.