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$$\Delta$ we equate $${\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,$$$$\lambda_+\lambda_-={\rm det}\,MM^t=(m_1\times m_2)^2=\|m_1\|^2\|m_2\|^2\sin^2\measuredangle \left( m_1, m_2 \right),$$ hence $$\lambda_\pm=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m_1\|^2+\|m_2\|^2\right)^2-\|m_1\|^2\|m_2\|^2\sin^2\measuredangle \left( m_1, m_2 \right)}.$$ If the two vectors $m_1$ and $m_2$ have the same norm $\|m\|$, this simplifies to $$\lambda_\pm=\|m\|^2\bigl(1\pm\cos \measuredangle \left( m_1, m_2 \right)\bigr),\;\;\text{if}\;\;\|m_1\|=\|m_2\|\equiv\|m\|.$$
This differs from the result in the OP. Let me check, as an example, $$M=\begin{pmatrix} 1&1\\ 0&1 \end{pmatrix},\;\;\sigma_1^2=\tfrac{3}{2}+\tfrac{1}{2}\sqrt 5,\;\;\sigma_2^2=\tfrac{3}{2}-\tfrac{1}{2}\sqrt 5.$$ Since the angle between the vectors $m_1={1\choose 0}$ and $m_2={1\choose 1}$ is $\pi/4$, the formula in the OP would give $\sigma_1^2=1+\sqrt 2$ and $\sigma_2^2=\sqrt 2$, which is incorrect.