Here are a few comments that you might find useful, though they don't completely solve the problem. First, using symmetries of the problem, you can easily reduce to the case that $f$ is mapping the interior of the unit circle $x^2+y^2<1$ diffeomorphically onto the interior of a circle $u^2+v^2 < r^2$ for some $r$. Second, because the Jacobian of the mapping is $\bigl|\sin\bigl(\phi(y)-\psi(x)\bigr)\bigr|\le 1$, it follows that $r\le 1$ with equality if and only if $\phi(y)-\psi(x)\equiv\pm\pi/2$, in which case, $\phi$ and $\psi$ would have to be constant, which is a trivial solution that has been ruled out. Hence, we can assume that $r<1$.
Let us assume that $f(x,y) = \bigl(u(x,y),v(x,y)\bigr)$ extends $C^2$ to the boundary circle $x^2+y^2=1$ (i.e., $\psi$ and $\phi$ extend differentiably to $[-1,1]$) and consider the curve
$\gamma(t) = f(\cos t ,\sin t)$, which maps the unit circle to the circle of radius $r<1$, which has curvature $1/r>1$. By the usual formula for the curvature of $\gamma$, the functions $\phi$ and $\psi$ must satisfy the first order relation
\begin{align}
(1/r)\bigl(1{-}2\,c\,s\,\cos(\psi(c){-}\phi(s))\bigr)^{3/2}
&= \sin(\phi(s){-}\psi(c))\\
&\qquad +(c\cos(\psi(c){-}\phi(s))-s)(1{-}c^2)\,\psi'(c)\\
&\qquad - (s\cos(\psi(c){-}\phi(s))-c)(1{-}s^2)\,\phi'(s)\\
\end{align}
where, to save writing, I am using $c = \cos t$ and $s=\sin t$.
However, this is a problem because, setting $t=\pi/2$, we get
$$
1/r = \sin(\phi(1)-\psi(0))-\psi'(0),
$$
while setting $t=-\pi/2$, we get
$$
1/r = \sin(\phi(-1)-\psi(0))+\psi'(0).
$$
Adding these two equations, we get
$$
2/r = \sin(\phi(1)-\psi(0)) + \sin(\phi(-1)-\psi(0)).
$$
But, since $r<1$, the left hand side of this equation is greater than $2$, while each of the terms on the right hand side have absolute value less than or equal to $1$. Impossible.
Thus, such an $f$, if it exists, cannot extend twice differentiablly to the boundary circle $x^2+y^2=1$.
Addendum: In the comments, the OP asked whether, if one dropped the assumption that the image of $f$ is a disk, but kept the assumption that $f$ extends $C^1$ to the boundary circle $x^2+y^2=1$, one could still show that the length of the boundary would be at most $2\pi$. I don't know how to do that in full generality, but I can show that, sufficiently near the 'trivial' case, where $\phi(y) \equiv \pi/2$ and $\psi(x)\equiv0$, one has this inequality.
The point is to start with the formula for the length of $\gamma$, which is
$$
L(\gamma) = \int_0^{2\pi} |\gamma'(t)|\,\mathrm{d}t
= \int_0^{2\pi} \bigl(1{-}\sin(2t)\,\cos(\psi(\cos t){-}\phi(\sin t))\bigr)^{1/2}\,\mathrm{d}t
$$
Write $\phi(y)=\pi/2+\kappa(y)$ and this becomes
$$
L(\gamma)
= \int_0^{2\pi} \bigl(1{-}\sin(2t)\,\sin(\psi(\cos t){-}\kappa(\sin t))\bigr)^{1/2}\,\mathrm{d}t.
$$
The key is to break this up into four integrals over the four quadrants. For $0\le t\le \pi/2$, set
$$
\begin{align}
a_1(t) &= \psi(\cos t)-\kappa(\sin t)\\
a_2(t) &= -\psi(\cos(\pi{-}t))+\kappa(\sin(\pi{-}t))
= -\psi(-\cos t) + \kappa(\sin t)\\
a_3(t) &= \psi(\cos(\pi{+}t))-\kappa(\sin(\pi{+}t))
= \psi(-\cos t) - \kappa(-\sin t)\\
a_4(t) &=-\psi(\cos(-t))+\kappa(\sin (-t))
= -\psi(\cos t)+\kappa(-\sin t)
\end{align}
$$
and note that $a_1+a_2+a_3+a_4=0$. Moreover, we find that
$$
L(\gamma) = \sum_{i=1}^4
\int_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin a_i(t)\bigr)^{1/2}\,\mathrm{d}t
$$
For a small parameter $\lambda$ consider the function
$$
f(\lambda) = \sum_{i=1}^4
\int_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin (\lambda a_i(t))\bigr)^{1/2}\,\mathrm{d}t.
$$
Because the sum of the $a_i$ vanishes, the Taylor series expansion of $f$ at $\lambda=0$ to second order takes the form
$$
f(\lambda)\simeq 2\pi
- \lambda^2\int_0^{\pi/2}\frac{\sin^2(2t)}{8}\left(\sum_{i=1}^4 a_i(t)^2\right)\,\mathrm{d}t.
$$
Clearly, $f$ has a strict local maximum at $\lambda=0$
unless the $a_i$ vanish, in which case $L(\gamma) = 2\pi$.
