Consider a vector $x \in \mathbb{R}^p$ and say we have $k$ matrices $A_i \in \mathbb{R}^{p \times n}$. Now consider a matrix $Y := \Big [ x^\top A_i \Big ]_{i=1}^k$ whereby we indicate that $Y \in \mathbb{R}^{k \times n}$ is a matrix s.t its $i^{th}-$row is the $n-$dimensional vector $x^\top A_i$.
Can we express/upper-bound the spectral norm of $Y$ in terms of $x$ and the $A_i$s?
Yes you can. One way is to rewrite $Y$ as the product
$$Y = (I_k \otimes x') \begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix}$$
Then you can use norm inequalities: $$\|Y\| = \left\|(I_k \otimes x') \begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix}\right\| \leq \|I_k \otimes x'\| \cdot \left\|\begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix}\right\|$$
For $\| I_k \otimes x'\|$ use the fact that the eigenvalues of the Kronecker product of matrices are the product of the eigenvalues of the matrices, so you can find the spectral norm as $$\| I_k \otimes x'\|^2 = \lambda_\max\left( I_k \otimes xx'\right).$$ Now note that $\lambda_\max(I_k)=1$, and that $\lambda_\max(xx') = x'x$, so you have that $$\| I_k \otimes x'\|^2 = x'x.$$
For the other matrix, you similarly have that $$\left\| \begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix} \right\|^2 = \lambda_\max\left(\begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix}' \begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix} \right)= \lambda_\max\left(\sum_{i=1}^k A_i'A_i\right) \leq \sum_{i=1}^k \lambda_\max(A_i'A_i).$$
Putting together the pieces, one bound on the norm is $$\|Y\|^2 \leq \sum_{i=1}^k \|A_i\|^2\|x\|^2$$ where $\|\cdot\|$ denotes both matrix or vector norms.
