I do not know if there is a name for this problem. However, we can view this in
terms of the usual operator norm.

Given $A \in \mathrm{Mat}_{n \times k}(\mathbb{R}), B \in \mathrm{Mat}_{m \times k}(\mathbb{R})$, consider for $x \in \mathbb{R}^n, y \in \mathbb{R}^m$:

\begin{align*}
\underset{\substack{ \lvert \lvert x \rvert \rvert = 1 \\ \lvert \lvert y \rvert \rvert = 1}}{\mathrm{sup}} \lvert \lvert Ax + By \rvert \rvert
&= \underset{\substack{ \lvert \lvert x \rvert \rvert \leq 1 \\ \lvert \lvert y \rvert \rvert \leq 1}}{\mathrm{sup}} \lvert \lvert Ax + By \rvert \rvert \\
&= \underset{\substack{ \lvert \lvert x \rvert \rvert \leq 1 \\ \lvert \lvert y \rvert \rvert \leq 1}}{\mathrm{sup}} \, \left \lvert  \left \lvert \begin{pmatrix}A & B \\ 0 & 0\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} \right \rvert  \right\rvert \\
&= \underset{\left \lvert \left \lvert \begin{pmatrix} x \\ y \end{pmatrix} \right \rvert \right \rvert_m = 1}{\mathrm{sup}} \left \lvert  \left \lvert \begin{pmatrix}A & B \\ 0 & 0\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} \right \rvert  \right\rvert,
\end{align*}
where $\left \lvert \left \lvert \begin{pmatrix} x \\ y \end{pmatrix} \right \rvert \right \rvert_m := \max(\lvert \lvert x \rvert \rvert, \lvert \lvert y \rvert \rvert)$ is the supremum norm from viewing $\mathbb{R}^{n+m}$ as the product $\mathbb{R}^n \times \mathbb{R}^m$. Observe then that the right-most
expression is the operator norm induced from $\lvert \lvert \cdot \rvert \rvert_m$ evaluated on $\left(\begin{smallmatrix}A & B \\ 0 & 0\end{smallmatrix}\right)$.