Here's how one can construct a specific example to illustrate what can happen:
First, recall from my answer to this question that, if you have a smooth map $f:D\to\mathbb{R}^2$ with constant positive singular values, then, letting $(e_1,e_2)$ be the orthonormal frame field on $D$ such that $\bigl(f'(e_1),f'(e_2)\bigr)$ are orthogonal with $|f'(e_i)|=\sigma_i$ with $0<\sigma_1<\sigma_2$, then the dual $1$-forms $(\omega_1,\omega_2)$ satisfy
$$
\mathrm{d}\omega_1 = \kappa_1\,\omega_1\wedge\omega_2\qquad
\mathrm{d}\omega_2 = \kappa_2\,\omega_2\wedge\omega_1\,\tag1
$$
where $\kappa_i:D\to\mathbb{R}$ is the curvature function of the flow lines of $e_i$. Moreover, we know that
$$
\mathrm{d}(\kappa_1\,\omega_1) = \mathrm{d}(\kappa_2\,\omega_2) = 0.\tag2
$$
Conversely, given a surface $S$ endowed with a coframing $\omega = (\omega_1,\omega_2)$ and functions $\kappa_1$ and $\kappa_2$ that satisfy equations (1) and (2), one can construct a corresponding $f:D\to\mathbb{R}^2$ with constant singular values $\sigma_i$.
Now, it's not hard to construct such data $(\omega_1,\omega_2,\kappa_1,\kappa_2)$ for which $\kappa_1$ and $\kappa_2$ vanish along some curves. For example, on a domain $S$ in the $uv$-plane that contains the origin, assume that $\lambda_1$ and $\lambda_2$ are nonzero functions that satisfy the linear hyperbolic system
$$
\frac{\partial\lambda_1}{\partial v} + u\lambda_2 =
\frac{\partial\lambda_2}{\partial u} + v\lambda_1 = 0\,.\tag3
$$
For example, $\lambda_1 = \lambda_2 = \mathrm{e}^{-uv}$ satisfies (3) on the entire $uv$-plane. Then the data
$$
\omega_1 = \lambda_1\,\mathrm{d}u,\quad
\omega_2 = \lambda_2\,\mathrm{d}v,\quad
\kappa_1 = \frac{u}{\lambda_1}\,,\quad
\kappa_2 = \frac{v}{\lambda_2}
$$
satisfy (1) and (2) while $\kappa_1$ vanishes along $u=0$ and $\kappa_2$ vanishes along $v=0$. Applying the integration method from the above-mentioned answer, one can now use this data to construct the desired $f$.
Taking the solution $\lambda_1=\lambda_2 = \mathrm{e}^{-uv}$ of (3), one finds that, letting $F:\mathbb{C}\to\mathbb{C}$ be the entire holomorphic function that satisfies $F(0)=0$ and $F'(z) = \mathrm{e}^{iz^2/2}$, and, for each real $\sigma>1$, letting $S_\sigma:\mathbb{C}\to\mathbb{C}$ satisfy $S_\sigma(x+iy)= x + i\sigma y$, and letting $D\subset\mathbb{C}$ be a domain containing $0\in\mathbb{C}$ on which $F$ has an inverse $F^{-1}:D\to\mathbb{C}$ satisfying $F^{-1}(0)=0$, then the mapping $f_\sigma:D\to\mathbb{C}$ defined by $f_\sigma(z) = F\bigl(S_\sigma\bigl(F^{-1}(z)\bigr)\bigr)$ has constant singular values $1$ and $\sigma$ while the corresponding functions $\kappa_1$ and $\kappa_2$ change sign along curves that meet transversely at the origin in $D$.
More generally, it is useful to switch points of view and let $F$ be the set of triples $(x;e_1,e_2)$ where $x\in\mathbb{R}^2$ and $(e_1,e_2)$ are an oriented orthonormal basis of $\mathbb{R}^2$. Then $F$ is a smooth $3$-manifold embedded naturally in $(\mathbb{R}^2)^3 = \mathbb{R}^6$.
Define the canonical $1$-forms $\omega_i = e_i\cdot \mathrm{d}x$ and $\omega_{12} = e_1\cdot\mathrm{d}e_2$, and note that they satisfy
$$
\mathrm{d}\omega_1 = -\omega_{12}\wedge\omega_2\,,\quad
\mathrm{d}\omega_2 = \omega_{12}\wedge\omega_1\,,\quad
\mathrm{d}\omega_{12} = 0.\quad
$$
Now, on $F\times\mathbb{R}^2$ with projection $(\kappa_1,\kappa_2):F\times\mathbb{R}^2\to\mathbb{R}^2$ onto the second factor, consider the exterior differential ideal $\mathcal{I}$ generated by the $1$-form $\theta = \omega_{12}+\kappa_1\,\omega_1 -\kappa_2\,\omega_2$ (which is a contact form on the $5$-manifold $F\times\mathbb{R}^2$) and the pair of $2$-forms
$$
\Upsilon_1 = (\mathrm{d}\kappa_1-{\kappa_1}^2\,\omega_2)\wedge\omega_1
\qquad
\Upsilon_2 = (\mathrm{d}\kappa_2-{\kappa_2}^2\,\omega_1)\wedge\omega_2
$$
Then $\mathrm{d}\theta \equiv \Upsilon_1 - \Upsilon_2 \mod \theta$. Then $\mathcal{I}$ is involutive for the independence condition $\omega_1\wedge\omega_2\not=0$.
In particular, each of the integral curves of the rank 4 system
$$
\theta = \mathrm{d}\kappa_1-\omega_1-{\kappa_1}^2\,\omega_2
= \mathrm{d}\kappa_2-\omega_2-{\kappa_2}^2\,\omega_1
= \omega_1-\omega_2 = 0
$$
on $F\times\mathbb{R}^2$ can be thickened to an integral surface of $\mathcal{I}$ on which $\omega_1\wedge\omega_2$ is nonvanishing. For example, we have
the integral curve
$$
(x;e_1,e_2,\kappa_1,\kappa_2) = \bigl((t,t);\,\partial/\partial x,\,\partial/\partial y, \,\arctan(t), \,\arctan(t)\bigr),
$$
which extends to an integral manifold of $\mathcal{I}$ on an open domain $D$ containing the origin $(0,0)$. The corresponding solution $f$ will have each of $\kappa_1$ and $\kappa_2$ changing sign along a curve that meets the 'diagonal' transversely. (One could probably solve this initial value problem explicitly, but I don't have time to attempt that right now.)
