This is also false. Here is a counterexample.

A = \begin{bmatrix}
 1 & 1/\sqrt{2}\\\\
 1/\sqrt{2} & 1
\end{bmatrix}

B = \begin{bmatrix}
 0 & -1/\sqrt{2}\\\\
 1/\sqrt{2} & 0
\end{bmatrix}

Then, the said block matrix has eigenvalues $(0,0,2,2)$, while

$\lambda^{1/2}(B^TB) = (1/\sqrt{2},1/\sqrt{2})$ and

$\lambda(A) = (1+1/\sqrt{2},1-1/\sqrt{2}))$