The singular value map is continuous and onto the ordered cone

Continuity. By the Courant-Fischer min-max formula,
$$\sigma_k(A)=\max_{\dim V=k}\ \min_{x\in V,|x|=1}|Ax|.$$
If $|Ax|\le|Bx|+\|A-B\|$ for all unit $x$, then taking the $\min$ over $x\in V$ and the $\max$ over $V$ gives $\sigma_k(A)\le\sigma_k(B)+\|A-B\|$. Exchanging $A$ and $B$,
$$|\sigma_k(A)-\sigma_k(B)|\le\|A-B\|.$$
So each $\sigma_k$ is $1$-Lipschitz, hence $\Sigma$ is continuous.

Image in $K$. By definition $\sigma_1\ge\dots\ge\sigma_n\ge0$.

Onto. Given $(s_1\ge\dots\ge s_n\ge0)\in K$, the matrix $D=\operatorname{diag}(s_1,\dots,s_n)$ has $D^TD=\operatorname{diag}(s_i^2)$, so its singular values are exactly $s_1,\dots,s_n$ in this order.

Remark. The fibres of $\Sigma$ are the orbits $\{U D V:U,V\in O(n)\}$ of the action of $O(n)\times O(n)$: two matrices have the same singular values iff they are orthogonally equivalent. So $K$ is the quotient $M_n(\mathbb{R})/(O(n)\times O(n))$.