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