There are many different answers to this question.
1. The open unit disc is just a model of the hyperbolic plane in which it turns out to be convenient to define a straight line in a certain way which does not coincide with the Euclidean definition. If it bothers you, you could put inverted commas around the word "straight" for a while, until you found that they made no difference.
2. If you do the calculations (given in the course) you find that the straight lines as defined in the course are the same as the geodesics , that is, the shortest curves between two points.
3. The reason straight lines in the disc model do not appear to be straight is that the hyperbolic plane is negatively curved (which means, roughly speaking, that the neighbourhood of every point is something like a Pringle crisp) so that it cannot be drawn in the Euclidean plane without distortion. This situation is rather like the impossibility of drawing a flat map of the world without distortion (this time because the sphere is positively curved). If you fly by aeroplane as directly as possible (that is, along a great circle) from London to San Francisco, and then trace your path on a wall atlas, given by Mercator's projection, say, then the curve you draw will be far from straight. In terms of the map itself (i.e., a certain model of the sphere) this is because distances are much smaller at the top and bottom of the map, so it pays to take an apparent detour, in this case to near the North Pole (which is represented by a line). An analogous situation occurs with the disc model, except that this time distances become larger near the edge, so that one takes a detour, which again isn't really a detour, to nearer the centre. A very similar argument applies to the half-plane model.
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