Using SVD to express the normalized norm of the output of a linear map This is probably trivial, but it's been quite some time since my Linear Algebra exam, and my SVD skills are rusty. 
1) Let $T$ be a linear map from $\mathbb{R}^n$ to $\mathbb{R}^n$, and $M$ the associated square matrix in the canonical basis.
$$\forall x\in \mathbb{R}^n,\ T(x)=Mx=U\Sigma Vx$$ 
I want to express 
$$\frac{||Mx||}{||x||}$$
in terms of the singular values of $M$ ($||x||\neq0$ obviously). In a presentation, I read that
$$\frac{||Mx||}{||x||}=\sqrt{\frac{\sum_i\sigma_i^2}{\sum_ix_i^2}}$$
where the $\sigma_i$ are the singular values of $M$. This is clearly wrong (e.g., consider an isotropic $M$). What's the correct expression? I would really like to see the passages of the proof, so that next time I can derive the correct expression myself. 
2) If $T$ is from $\mathbb{R}^m$ to $\mathbb{R}^n$, i.e., $M$ is not square, is it still possible to express $\frac{||Mx||}{||x||}$ in terms of the singular values of $M$?
 A: Let $\{\sigma_1,\dots,\sigma_r\}$ be the nonzero singular values of $M$ (where $r$ is the rank of $M$). As you pointed out (assuming the matrix $M$ has real entries) there exists an $m\times m$ orthogonal matrix $U$, an $n\times n$ orthogonal matrix $V$, and a diagonal $m\times n$ matrix $\Sigma$ (whose nonzero diagonal entries are the singular values $\sigma_1,\dots,\sigma_r$) such that 
$$
M=U\Sigma V.
$$
Let $\{v_1,\dots,v_m\}$ denote the rows of $V$ and $\{u_1,\dots,u_n\}$  the columns of $V$ (both of which are orthonormal bases). In particular, it holds that $v_i^Tv_j = 0 $ for indices $i\neq j$ and $v_i^Tv_i=1$ for all $i$. (An analogous statement holds for the vectors $u_i$.)
We can expand $x$ in the basis $\{v_1,\dots,v_m\}$  as 
$$
x= x_1v_1+\cdots+x_nv_n
$$
where $x_1,\dots,x_n\in\mathbb{R}$ are the coefficients of the vector $x$ in this basis such that 
$$
\lVert x\rVert = \sqrt{\sum_{i=1}^n x_i^2}.
$$
Multiplying out, we find that $Mx$ expanded out in the $\{u_1,\dots,u_m\}$ basis is
$$
Mx = \sigma_1 x_1 u_1 + \cdots + \sigma_r x_r u_r.
$$
such that $\lVert Mx\rVert^2 = \sigma_1^2 x_1^2+\cdots+ \sigma_r^2 x_r^2$. I don't think you can really say much else.
To address your second question, note that
$$
\sup_{x\neq 0} \frac{\lVert Mx\rVert}{\lVert x\rVert} = \sigma_1,
$$
which is the largest singular value. Indeed, assuming $\sigma_1\geq \sigma_i$ for all other $i\in\{2,\dots,r\}$, we see that
$$
\lVert Mx\rVert =\sqrt{ \sum_{i=1}^r \sigma_i^2 x_i^2} \leq \sigma_1\sqrt{\sum_{i=1}^n x_i^2}  = \sigma_1 \lVert x\rVert,
$$
with equality if and only if $x$ is parallel to $v_1$ (i.e. $x_i=0$ for all $i\in\{2,\dots,n\}$).

We can write $U$ and $V$ as
$$
U = \sum_{i=1}^m u_ie_i^T\qquad\text{and }\qquad V = \sum_{i=1}^n e_iv_i^T,
$$
where the $e_i$'s are the "standard" basis vectors (i.e., with zeros everywhere except a 1 in the $i$th position), and $\Sigma$ as
$$
\Sigma = \sum_{i=1}^r \sigma_i \, e_ie_i^T.
$$
It follows that $Vv_i = e_i$ and $Ue_i = u_i$ for all $i$. Moreover, $\Sigma e_i = \sigma_i e_i$. Thus
$$
Mv_i = U\Sigma Vv_i = U\Sigma e_i = \sigma_i Ue_i = \sigma_i u_i.
$$
