$$\begin{pmatrix} v^\dagger \\ U \end{pmatrix}\begin{pmatrix} v & U^\dagger \end{pmatrix} = \begin{pmatrix}v^\dagger v & v^\dagger U^\dagger \\ U v & UU^\dagger\end{pmatrix} = \mathbf{1}_n,$$
so
$v^\dagger v = 1$ and $Uv = 0$ and $U\in\mathrm{U}(n-1)$. We get a map from this to $\mathrm{SU}(n)$ by $(v,U)\mapsto \begin{pmatrix} \mathrm{det}(U)^{-1} & 0 \\ 0 & U \end{pmatrix}$ in a basis where $v$ is the first basis vector.