It's a model of hyperbolic geometry in the plane, the same geometry described by Lobachevsky. There are several models of planar hyperbolic geometry. The most common ones include the Poincaré disk (or more generally ball), the Poincaré upper hald-plane (or more generally half-space), the Beltrami-Klein model which is somtimes also called the projective model, and the hyperboloid model which uses a three-dimensional Minkowsky space to embed the plane.
Comparing these models, the Poincaré disk model has the benefit that it doesn't extend to infinity (as the upper half-plane does), and preserves angles (contrary to the non-Poincaré models), and lives in the plane without need for a third dimension (as needed by the hyperboloid). But these benefits are balanced by drawbacks: the upper half-plane can use real $2\times2$ matrices to describe isometric transformations, the Beltrami-Klein model uses straight lines to model geodesics, and the hyperboloid model is very close in formulation to the geometry on a sphere.
An imperfect model of this geometry would – at least in my interpretation of the word “imperfect” – be the geometry on the tractricoid. Since the tractricoid has constant negative curvature, it closely resembles the hyperbolic plane, uisng “real” angles and “real” geodesics, as they are in the ambient three-space. But the tractricoid only models a part of the hyperbolic plane; it contains closed curves which would not close in the hyperbolic plane. So it is only a local model, which doesn't represent the global structure well. There can be no embedding of the hyperbolic plane into real three-space which uses normal Euclidean angles and geodesics.
You could define all kinds of stuff about what you call a “line”, what you call “distance” and so on. The special thing about hyperbolic geometry is the fact that it still satisfies the first four axioms of Euclid, even though it violates the fifth. In this regard it is pretty much unique.
Best Answer
The easiest is probably to write it down in complex coordinates. A unit-speed geodesic passing through the origin $0\in{\Bbb D}$ at time zero in the direction of $n\in {\Bbb C}$ with $|n|=1$ is given by $$ w_{n}(t) = n \tanh \frac{t}{2}.$$ It is a straight line-segment keeping the angle $\theta$ with the real axis. Now, given $p,q\in{\Bbb D}$, the Möbius transformation: $$ w=R(z) = \frac{z-p}{1-\bar{p}z} \Leftrightarrow z=R^{-1}(w) = \frac{w+p}{1+\bar{p}w}$$ is an isometry that maps $p$ to 0. Let $q'=R(q)=\frac{q-p}{1-\bar{p}q}$ and set $n=q'/|q'|$. To get the unit speed geodesic trough $p$ and $q$ (passing $p$ at time zero) simply take the preimage of $w_n(t)$: $$ z(t) = R^{-1}(w_n(t)).$$ You may here insert the various expressions to get a rather messy looking explicit formula (it becomes a fractional linear map in $\tanh(t/2)$ but with rather complicated coefficients). Depending on what you want to do with the geodesic you might not want to use this explicit formula.
Möbius transformations preserves the collection of disks union lines as well as angles (being conformal) so as you mention the trajectory of $z(t)$ must be part of a circle perpendicular to the unit circle.