When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $e^{\pm y}$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $x,y,\theta$ coordinates anymore, but in the coframing. When you write out the velocity of $\sigma(t)$ in the coframing, it doesn't turn out to be as straightforward. In a complex notation, the Cartan connection is simpler. On the plane, let $\underline{\xi}=e^{-y}dx+idy$ and $\underline{\gamma}=ie^{-y}dx$. Add a variable $\theta$ and let $\xi=e^{i\theta}\underline{\xi}$, and $\gamma=\underline{\gamma}+id\theta$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $e^{-y}$ factors. The 1-form $\gamma$ does not vanish on your curve $\sigma(t)$. So is not a geodesic. Your first curve $(x(t),y(t),\theta(t))=(0,t,0)$ is, since $\gamma=0$ on it.

Expanding out in these coordinates, the equation of a geodesic is $\xi=c \, ds, \gamma=0$, for a complex constant $c$. We can assume that $c\ne 0$. The 1-forms $\xi$ and $\gamma$ transform when we rotate $\theta$, say the action $r_{\theta_0} : \theta\mapsto \theta+\theta_0$, as $r_{\theta_0}^*\xi= e^{i\theta_0}\xi$, $r_{\theta_0}^*\gamma=\gamma$. So we change the equations by rotation of $c$, and we can arrange that $c>0$. Rescale the $s$ variable to get $c=1$. So our equations of geodesics expand out to $$ \cos \theta e^{-y} \frac{dx}{ds} - \sin \theta \frac{dy}{ds}=1, $$ $$ \sin \theta e^{-y} \frac{dx}{ds} + \cos \theta \frac{dy}{ds}=0, $$ $$ \frac{d\theta}{ds}+e^{-y}\frac{dx}{ds}=0, $$ if I have calculated correctly.

The geodesics of Euclidean geometry in the plane are straight lines, i.e. curves that do not change direction. Their Cartan geometry is $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane, where $\xi=e^{i\theta}(dx+i \, dy)$, $\gamma=i \, d\theta$. So $\gamma=0$ says that the direction does not change: a straight line, i.e. a geodesic.

Now consider the hyperbolic plane. (When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $e^{\pm y}$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $x,y,\theta$ coordinates anymore, but in the coframing. When you write out the velocity of $\sigma(t)$ in the coframing, it doesn't turn out to be as straightforward.)

In a complex notation, the Cartan connection is simpler. On the plane, let $\underline{\xi}=e^{-y}dx+idy$ and $\underline{\gamma}=ie^{-y}dx$. We see this because we want metric $\xi_1^2+\xi_2^2$. Add a variable $\theta$ and let $\xi=e^{i\theta}\underline{\xi}$, and $\gamma=\underline{\gamma}+id\theta$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $e^{-y}$ factors. The 1-form $\gamma$ does not vanish on your curve $\sigma(t)$. So is not a geodesic. Your first curve $(x(t),y(t),\theta(t))=(0,t,0)$ is, since $\gamma=0$ on it.

Expanding out in these coordinates, the equation of a geodesic is $\xi=c \, ds, \gamma=0$, for a complex constant $c$. We can assume that $c\ne 0$. The 1-forms $\xi$ and $\gamma$ transform when we rotate $\theta$, say the action $r_{\theta_0} : \theta\mapsto \theta+\theta_0$, as $r_{\theta_0}^*\xi= e^{i\theta_0}\xi$, $r_{\theta_0}^*\gamma=\gamma$. So we change the equations by rotation of $c$, and we can arrange that $c>0$. Rescale the $s$ variable to get $c=1$. So our equations of geodesics expand out to $$ \cos \theta e^{-y} \frac{dx}{ds} - \sin \theta \frac{dy}{ds}=1, $$ $$ \sin \theta e^{-y} \frac{dx}{ds} + \cos \theta \frac{dy}{ds}=0, $$ $$ \frac{d\theta}{ds}+e^{-y}\frac{dx}{ds}=0, $$ if I have calculated correctly.

When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $e^{\pm y}$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $x,y,\theta$ coordinates anymore, but in the coframing. When you write out the velocity of $\sigma(t)$ in the coframing, it doesn't turn out to be as straightforward. In a complex notation, the Cartan connection is simpler. On the plane, let $\underline{\xi}=e^{-y}dx+idy$ and $\underline{\gamma}=ie^{-y}dx$. Add a variable $\theta$ and let $\xi=e^{i\theta}\underline{\xi}$, and $\gamma=\underline{\gamma}+id\theta$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $e^{-y}$ factors. The 1-form $\gamma$ does not vanish on your curve $\sigma(t)$. So is not a geodesic. Your first curve $(x(t),y(t),\theta(t))=(0,t,0)$ is, since $\gamma=0$ on it.

When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $e^{\pm y}$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $x,y,\theta$ coordinates anymore, but in the coframing. When you write out the velocity of $\sigma(t)$ in the coframing, it doesn't turn out to be as straightforward. In a complex notation, the Cartan connection is simpler. On the plane, let $\underline{\xi}=e^{-y}dx+idy$ and $\underline{\gamma}=ie^{-y}dx$. Add a variable $\theta$ and let $\xi=e^{i\theta}\underline{\xi}$, and $\gamma=\underline{\gamma}+id\theta$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $e^{-y}$ factors. The 1-form $\gamma$ does not vanish on your curve $\sigma(t)$. So is not a geodesic. Your first curve $(x(t),y(t),\theta(t))=(0,t,0)$ is, since $\gamma=0$ on it.

Expanding out in these coordinates, the equation of a geodesic is $\xi=c \, ds, \gamma=0$, for a complex constant $c$. We can assume that $c\ne 0$. The 1-forms $\xi$ and $\gamma$ transform when we rotate $\theta$, say the action $r_{\theta_0} : \theta\mapsto \theta+\theta_0$, as $r_{\theta_0}^*\xi= e^{i\theta_0}\xi$, $r_{\theta_0}^*\gamma=\gamma$. So we change the equations by rotation of $c$, and we can arrange that $c>0$. Rescale the $s$ variable to get $c=1$. So our equations of geodesics expand out to $$ \cos \theta e^{-y} \frac{dx}{ds} - \sin \theta \frac{dy}{ds}=1, $$ $$ \sin \theta e^{-y} \frac{dx}{ds} + \cos \theta \frac{dy}{ds}=0, $$ $$ \frac{d\theta}{ds}+e^{-y}\frac{dx}{ds}=0, $$ if I have calculated correctly.