Observe that
$M=\begin{pmatrix}U_{11}\Phi & U_{12}\Phi \\ U_{21}\Phi & U_{22}\Phi\end{pmatrix}=\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}$.
Then by property $\det(AB)=\det(A)\det(B)$, the determinant of $M$ is
$\det(M)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)\det\left(\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}\right)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)[\det(\Phi)]^2$,
where the last step comes from the fact that the determinant of a block diagonal matrix is the product of the determinants of the diagonal blocks of such matrix (see General expression for determinant of a block-diagonal matrix).