If $a_{1},a_{2} \in \mathcal{R}^{2},$ $\ \|a_{1}\|_{2} = \|a_{2}\|_{2} = K$, and the angle $\theta$ between $a_{1}$ and $a_{2}$ is between $0$ and $\pi/2$, we want to compute the SVD of the matrix $A = [a_{1},a_{2}],$ i.e. the vectors $a_{1}, a_{2}$ are the columns of $A$.
We know that $a_{2} = Ra_{1}$, where $R$ is a rotation matrix. If $Pa_{2} = a_{1}$, then $(I-P)a_{2}$ is orthogonal to $a_{1}$, where $P = R^{-1}$. But I am not sure how to use these to form an SVD of $A$...
We seek the SVD $A = U\Sigma V^*$. Note that $$ A^*A = U(\Sigma\Sigma^*) U^* = \pmatrix {\langle a_1,a_1 \rangle & \langle a_2,a_1 \rangle\\ \langle a_1,a_2 \rangle & \langle a_2,a_2 \rangle } = K^2\pmatrix {1 & \cos\theta\\ \cos \theta & 1 } $$ This allows you to find $U$ and $\Sigma$ (by finding a spectral decomposition of the above). From there, note that $$ V = A^{-1} U \Sigma $$ Or, the easier to compute $$ V = [\Sigma^{-1}U^*A]^* $$ To be sure, your singular values will be $\sigma_1 = \sqrt{1 + \cos \theta}$ and $\sigma_2 = \sqrt{1 - \cos \theta}$, and $$ U = \frac 1{\sqrt 2} \pmatrix{ 1 & 1\\ 1 & -1 } $$