Since $\sigma_1^2=\lambda_+$ and $\sigma_2^2=\lambda_-$ are the two eigenvalues of the symmetric matrix product $MM^t$, we have $\lambda_++\lambda_-={\rm tr}\,MM^t=\|m_1\|^2+\|m_2\|^2$. Hence we may write WLOG
$$\lambda_\pm=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)\pm\Delta.$$
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
$$\lambda_+\lambda_-=\tfrac{1}{4}\left(\|m_1\|^2+\|m_2\|^2\right)^2-\Delta^2,$$
hence
$$\lambda_\pm=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m_1\|^2+\|m_2\|^2\right)^2-\|m_1\|^2\|m_2\|^2\sin^2\measuredangle \left( m_1, m_2 \right)}$$
$$\qquad=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m_1\|^2-\|m_2\|^2\right)^2+\|m_1\|^2\|m_2\|^2\cos^2\measuredangle \left( m_1, m_2 \right)}.$$
I checked that this expression agrees with a direct calculation of the eigenvalues of $MM^t$, so I'm confident it is correct.    
 It seems to differ from the result in the OP, but perhaps it is the same, let me check further.