The short SVD of $uu^*$ is $v\sigma v^*$ with $v=\frac{u}{\|u\|}$ and $σ=\|u\|^2$.
If you compute the bisector $w=u+\frac{u_1+0}{|u_1|+0}\|u\|e_1$ of $u$ and $e_1$, then you can get a full unitary matrix $V$ as the reflection matrix $I-2\frac{ww^*}{\|w\|^2}$, so that $uu^*=V\Sigma V^*$
The singular value decomposition is obtained from the polar decomposition, together with the spectral theorem.
The polar decomposition gives you $A=V|A|$, where $V$ is a partial isometry such that $\operatorname{ran}V^*V=\overline{\operatorname{ran}A^*}$, and $|A|=(A^*A)^{1/2}$. Since $|A|\in B(H_1)$ and positive and compact, we apply the Spectral Theorem to obtain
$$\tag1
|A|=\sum_{j=1}^\infty\sigma_j\,P_j,
$$
where $\sigma_1\geq\sigma_2\geq\cdots\geq0$ and each $P_j$ is a rank-one projection. We can rewrite $(1)$ as
$$\tag2
|A|=U^*DU,
$$
where $U$ is a unitary and the is the diagonal operator (in the canonical basis, say) with diagonal $\sigma_1,\sigma_2,\ldots$
Then
$$\tag3
A=VU^*DU=WDU
$$
where $D$ is as above, $W$ is a partial isometry, and $U$ is unitary.
An often more useful way of writing this is choosing unit vectors $e_j$ with $P_je_j=e_j$ (so they form an orthonormal basis of the range of $|A|$) and write $(1)$ as
$$\tag4
|A|=\sum_{k=1}^\infty\sigma_k\,\langle\cdot,e_k\rangle \,e_k.
$$
Then
$$\tag5
A=V|A|=\sum_{k=1}^\infty\sigma_k\,\langle\cdot,e_k\rangle \,Ve_k.
$$
As $V$ is an isometry on $\operatorname{ran}|A|$, we get that $\{Ve_k\}$ is orthonormal. So the Singular Value Decomposition can be restated as saying
If $A\in L(H_1,H_2)$ is compact there exist orthonormal families $\{e_k\}\subset H_1$ and $\{f_k\}\subset H_2$ such that
$$\tag6
A=\sum_{k=1}^\infty\sigma_k\,\langle\cdot,e_k\rangle \,f_k.
$$
Best Answer
You have $$\tag1 AA^*=U\Sigma^2 U^*.$$ Taking the trace in $(1)$, $$ \operatorname{Tr}(\Sigma^2)=\operatorname{Tr}(AA^*)=\operatorname{Tr}(U\Sigma^2U^*)=\operatorname{Tr}(\Sigma^2 U^*U). $$ Then $$ 0=\operatorname{Tr}(\Sigma^2\,(I-U^*U))=\operatorname{Tr}(\Sigma\,(I-U^*U)^2\,\Sigma) $$ As the trace is faithful, we get $\Sigma(I-U^*U)^2\Sigma=0$, and so $(I-U^*U)\Sigma=0$. So $$\tag2 \Sigma=U^*U\Sigma=\Sigma\,U^*U. $$ Now $$ A^*A=V\Sigma U^*U\Sigma V^*=V\Sigma^2\,V^*. $$ Now $$ \ker A=\ker A^*A=\ker V\Sigma^2 V^*=\ker \Sigma V^*. $$ So, taking orthogonals,
$$ \operatorname{ran} A^*=\operatorname{ran}V\Sigma. $$ This shows that the first $r$ columns of $V$ span the range of $A^*$ (which is the same as the range of $|A|$). Going back to $(1)$, $$ \ker A^*=\ker AA^*=\ker U\Sigma^2\,U^*=\ker \Sigma U^*, $$ so $$ \operatorname{ran} A=\operatorname{ran} U\Sigma, $$ so the first $r$ columns of $U$ span the range of $A$.
It is not true in general that $\ker A=\ker U$. For instance take $$ A=\begin{bmatrix} 0&0\\1&0\end{bmatrix} \,\begin{bmatrix} 1&0\\0&0\end{bmatrix} \,\begin{bmatrix} 0&1\\1&0\end{bmatrix} =\begin{bmatrix} 0&0\\0&1\end{bmatrix}. $$