In one of the eight Thurston geometries there is the geometry of the universal cover of $SL(2, \mathbb{R})$. But from the algebraic point of view $PSL(2,\mathbb{R})$ is sufficient for building 3-manifolds i.e. we let the group act on itself and then quotient out a discrete subgroup of it. This is what we do for the nilgeometry where we use the 2-step nilpotent Heisenberg group instead of $PSL(2,\mathbb{R})$.

Thus my question is, why passing to the universal cover? Professor told me that it is for obtaining a simply-connected 3-manifold so as to include as many manifolds as possible for the classification i.e. there is a preference for starting with simply-connected manifolds. But I do not really understand the reason behind it although I guess there should be a quick answer to it.

Let me try to clarify my question a bit. My intention is to make a comparison between nilgeometry and $SL(2,\mathbb{R})$-geometry. Their algebraic construction is the same except that one starts with the 2-step nilpotent Heisenberg group and the other with $PSL(2,\mathbb{R})$. However for $SL(2,\mathbb{R})$-geometry there is an additional step of passing to the universal cover of $SL(2,\mathbb{R})$. What I am asking is the motivation for this extra step.