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Since $\sigma_1^2$ and $\sigma_2^2$ are the two eigenvalues of the symmetric matrix product $MM^t$, we have $\sigma_1^2+\sigma_2^2={\rm tr}\,MM^t=\|m_1\|^2+\|m_2\|^2$. Hence we may write WLOG $$\sigma_1^2=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)+\Delta,\;\;\sigma_1^2=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)-\Delta.$$ To determine $\Delta>0$ we equate $${\rm det}\,MM^t=(m_1\times m_2)^2=\|m_1\|^2\|m_2\|^2\sin^2\measuredangle \left( m_1, m_2 \right)$$ to $$\sigma_1^2\sigma_2^2=\tfrac{1}{4}\left(\|m_1\|^2+\|m_2\|^2\right)^2-\Delta^2,$$ and the formula in the OP follows.