Show that $\gamma = \{r = const >0\}$ is NOT a geodesic on the plane with parametrisation $ds^2 = dr^2 + \sinh^2 d\phi^2$ of the hyperbolic plane.
I know how to work this out in principle. My confusion is with the notation $\gamma = \{r = const >0\}$. Should I take this to mean $\gamma(t) = (r, t)$?
Otherwise, what $\gamma(t)$ do I need to use in the geodesic equation: $\nabla_\gamma \dot \gamma =0$?
You need a parametrization of the curve, which is a circle of radius $r$. In polar coordinates you can, in fact, use $t\mapsto (r(t), \phi(t)) = (r, t)$ (or $(r, ct)$ with a normalizing factor $c$ to achieve a unique speed parametrization).
Alternatively, in the standard Euclidean coordinates, you may use $(r\sin(ct), r\cos(ct))$, but with the metric you are using polar coordinates will be better suited.