This is a self-answered question, which I post since it wasn't immediately trivial for me. Of course, I would be happy to find easier or quicker solutions.
Let $A:\mathbb{R}^2 \to \mathbb{R}^2$ be an invertible linear map, and let $v_1,v_2$ be an orthonormal basis for $\mathbb{R}^2$. Suppose that you are given the following three numbers $c_1,c_2,c_3$:
$$ c_i=|Av_i|, c_3=\langle Av_1,Av_2 \rangle$$
$c_1,c_2,c_3$ determine $A$ up to a left multiplication by an orthogonal matrix (see details below). In particular, they determine the singular values $\sigma_i$ of $A$. Is there a way to compute the $\sigma_i$ explicitly in terms of the $c_i$? Can we find a closed-form formula?
I am fine with assuming that $\det A$ is positive.
Comment: This makes sense by from the perspective of counting degrees of freedom: $M_2(\mathbb{R})$ has dimension $4$, and we are given $3$ constraints-so we expect to be left with one (continuous) degree of freedom-here corresponds to the angle of a rotation in $\text{SO}(2)$. (We also have one discrete degree of freedom of a possible reflection, since the $c_i$ are indifferent to that).
Note that $|\det A|$ is equal to the area of the parallelogram whose edges are $Av_1,Av_2$- i.e. $|\det A|^2=|Av_1 \times Av_2|^2=|Av_1|^2|Av_2|^2-(\langle Av_1,Av_2 \rangle)^2=c_1^2c_2^2-c_3^2$. So,
$$ \sigma_1 \sigma_2=|\det A|=\sqrt{c_1^2c_2^2-c_3^2}. \tag{1}$$
A proof that the $c_i$ determine $A$ up to a left orthogonal multiplication:
Suppose that $A,B$ are linear invertible maps with the same $c_i$. Write $S=AB^{-1}$. Set $w_i=Bv_i$. Then $Sw_i=Av_i$, so $|Sw_i|=|Av_i|=|Bv_i|=|w_i|$. Furthermore, $\langle Sw_1,Sw_2 \rangle=\langle Av_1,Av_2 \rangle=\langle Bv_1,Bv_2 \rangle=\langle w_1,w_2 \rangle$. Thus $S$ is an isometry.
(Note that the $w_i$ form a basis, since $B$ is invertible).
Alternative way to see that $A$ is uniquely determined up to an isometry, is to see that $B=A^TA$ is uniquely determined by the $c_i$. (Thinking of the polar decomposition of $A$, $A=OP$, where $O$ is orthogonal, and $P$ is symmetric positive-definite, we have $P^2=B$. Determining $A$ up to an isometry is equivalent to determining the positive factor $P$). Indeed, it is immediate from the definitions that
$$ \langle Bv_1,v_1 \rangle=c_1^2, \langle Bv_1,v_2 \rangle=\langle Bv_2,v_1 \rangle=c_3, \langle Bv_2,v_2 \rangle=c_2^2.$$
This information determines $B$ uniquely. In fact, we can use $B$ to recover $A$ itself up to isometry: $\sqrt{B}$ is a representative for $A$, i.e. $A =Q \sqrt{B}$, where $Q$ is an isometry.
Representing $B$ w.r.t the basis $v_i$, we have $$B=\begin{pmatrix} c_1^2 & c_3 \\\ c_3 & c_2^2 \end{pmatrix}.$$
Another point of view is that the $c_i$ determine $\langle Av,Aw \rangle$ for all $v,w \in \mathbb{R}^2$ (seen via expanding $v,w$ in terms of the basis $v_i$). Since $\langle Av,Aw \rangle=\langle A^TAv,w \rangle$ this is equivalent to determining $A^TA$.
Consider $B=A^TA$-we are interested in the eigenvalues of $B$, and we know the following three quantities (in terms of the $c_i$):
$$ \langle Bv_1,v_1 \rangle=c_1^2, \langle Bv_1,v_2 \rangle=\langle Bv_2,v_1 \rangle=c_3, \langle Bv_2,v_2 \rangle=c_2^2.$$
Write $T_{\lambda}=B-\lambda \text{Id}$. We look for $\lambda$ satisfying $\det(T_{\lambda})=0$ or equivalently $T_{\lambda}v_1 \wedge T_{\lambda}v_2=0$.
We have $$ \langle T_{\lambda}v_1,v_1 \rangle=c_1^2-\lambda, \langle T_{\lambda}v_1,v_2 \rangle=\langle T_{\lambda}v_2,v_1 \rangle=c_3, \langle T_{\lambda}v_2,v_2 \rangle=c_2^2-\lambda,$$
which imply (since the $v_i$ are orthonormal) that
$$ T_{\lambda}v_1=(c_1^2-\lambda)v_1+c_3v_2, T_{\lambda}v_2=c_3v_1+(c_2^2-\lambda)v_2.$$
So, we obtain
$$ 0=T_{\lambda}v_1 \wedge T_{\lambda}v_2=\big((c_1^2-\lambda)(c_2^2-\lambda)-c_3^2\big)v_1 \wedge v_2. $$
To, conclude we have $$ (c_1^2-\lambda)(c_2^2-\lambda)-c_3^2=0, \tag{2} $$
or $$ \lambda^2-\lambda(c_1^2+c_2^2)+(c_1^2c_2^2-c_3^2)=0, $$
so
$$ \sigma_{1,2}^2=\lambda_{1,2}=\frac{c_1^2+c_2^2 \pm \sqrt{(c_1^2-c_2^2)^2+4c_3^2}}{2}. $$
Note that when $c_3=0$ ($Av_1,Av_2$ are orthogonal) we get $\sigma_{1,2}=c_1,c_2$ as expected.
BTW, equation $(2)$ is similar to equation $(1)$ in the question-think of $\lambda=0$. Perhaps there is a more immediate way to deduce it.