Given a vector $y$ defined as the product of a (non-square) matrix $A$ and vector $x$ i.e. $Ax$, such that matrix $A \in \mathbb{R}^{(m, n)}$ and $\lvert\lvert x \rvert \rvert \leq C$, how can we find map $Y$ as bounding value to vector $y$ i.e. $\lvert\lvert y \rvert\rvert$?
I find this notion of spectral radius $\lvert\lvert A \, x \rvert\rvert \leq \rho(A) \, \lvert\lvert x \rvert\rvert$ mostly for squared matrices. In this reference [1], they use the square root of the greatest singular value of matrix $A A^\intercal$
The same definition works for non-square matrices.
$$\|A\|:=\sup_{\|x\| = 1} \|Ax\|=\sup_{x^\top x=1} \sqrt{x^\top A^\top A\, x}=\max\{\sqrt{\lambda}:\lambda\text{ is an eigenvalue of }A^\top A\}$$
Here $x\in\Bbb{R}^n, A\in M_{m\times n}(\Bbb{R}),Ax\in\Bbb{R}^m$.
The matrices $A^\top A$ and and $A A^\top$ have the same positive eigenvalues (including multiplicity). This follows from the singular value decomposition. They're both symmetric and non-negative so their eigenvalues are $\ge 0$. This holds whether $A$ is square or not.
Over $\Bbb{C}$ use $A^\dagger$ instead of $A^\top$.