Seifert manifolds and Fuchsian group A Fuchsian group is a discrete subgroup of $PSL(2, \mathbb{R})$. 
Let $M$ be a Seifert manifold (maybe with boundary) and $t \in \pi_1(M)$ the class of a regular Seifert fiber. 
Hempel claims in his books on 3-Manifolds that the group $G = \pi_1(M)/ \langle t \rangle$ is a
Fuchsian group.
Q: How can one see this? I can kind of imagine this when the Seifert surface is a hyperbolic surface.
To answer this question it might be helpful to know that the group $G$ has the following representation.
$$G = \langle a_1, b_1,\ldots, a_g, b_g, c_1, \ldots, c_q, d_1, \ldots, d_k : c_i^{n_i} = 1,  [a_1, b_1]\ldots[a_g,b_g]c_1\ldots c_q d_1 \ldots d_k =1 \rangle $$
Edit: One definitly has to make the claim a bit more precise. I definitly don't see this with M = S^3. 
 A: This claim is, strictly speaking, false, my guess is that you are missing some hypothesis. All Seifert manifolds fall into 3 main classes: The ones with spherical base-orbifold, ones with Euclidean base-orbifolds and ones with hyperbolic base, orbifolds. You can find a detailed discussion of this in Peter Scott "Geometries of 3-manifolds". In all cases, the quotient group $G$ in your question is the fundamental group of the base-orbifold. If the base is hyperbolic, then $G$ embeds as a discrete subgroup in $SO(2,1)$, the full isometry group of the hyperbolic plane. The group $PSL(2,R)$ is surely not enough since it preserves the orientation on the hyperbolic plane, but you can just take a circle bundle over a nonorientable hyperbolic surface, as your Seifert manifold.
In the case of Euclidean base, the group $G$ is never Fuchsian as it contains a rank 2 free abelian subgroup. In the case of spherical base, sometimes the group $G$ is Fuchsian, i.e., when it is cyclic or dihedral. But in some cases it is not Fuchsian, for instance, when it is isomorphic to $A_4$, the alternating group on 4 letters. The reason is that all finite subgroups of $SO(2,1)$ are cyclic or dihedral and $A_4$ is not.     
The correct statement is that the group $G$ acts as a discrete isometry group either on the 2-dimensional sphere or on the Euclidean plane or on the hyperbolic plane, take a look at Scott's paper. 
