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
$$\lambda_\pm=\tfrac{1}{2}\left(\|m_1\|^2+\|m_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m_1\|^2+\|m_2\|^2\right)^2-\|m_1\|^2\|m_2\|^2\sin^2\measuredangle \left( m_1, m_2 \right)}.$$
I checked that this expression agrees with a direct calculation of the eigenvalues of $MM^t$, so I'm confident it is correct.    

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 It differs from the result in the OP. Let me check the simple case 
$$M=\begin{pmatrix}
1&1\\
0&1
\end{pmatrix},\;\;\sigma_1^2=\tfrac{3}{2}+\tfrac{1}{2}\sqrt 5,\;\;\sigma_2^2=\tfrac{3}{2}-\tfrac{1}{2}\sqrt 5.$$
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.
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