Complement of figure 8 knot - zero vertex Thurston in "3-dimensional Geometry and Topology" explicitely creates the hyperbolic complement of the figure 8 knot by glueing two ideal tetrahedra.
What I do not understand is that he pushes the knot "to the vertex".
Isn't the vertex 0 dimensional (just a point). Could someone clarify how is this then the complement if the knot is a point? For me it seems like it should not be  a point. Does it get contracted to a single point?
 A: An ideal tetrahedron has no vertices.  The zero-skeleton (of Thurston's triangulation of the figure eight knot complement) is empty.
A: Yes, the knot is contracted to a single point.
If we glue two (compact, non-ideal) tetrahedra as described, we obtain a topological space $\overline{M}$ that is not a manifold. All vertices become identified to a singular point. But if we remove an open neighborhood $U$ of this point (say, $\varepsilon$-neighborhood with respect to the Euclidean metric on the tetrahedra), the result $\overline{M} \setminus U$ is the complement to a tubular neighborhood of the figure 8 knot. In particular, $\partial U$ is a torus. (The latter is easy to see: the links of the vertices of tetrahedra are glued so that there are six triangles around every vertex; by the Euler formula this is a surface of genus $0$. Remains to check that it is orientable.) More difficult is to understand the topology of $\overline{M} \setminus U$, see e.g. the diagram referred to in a comment to your question.
To summarize, $\overline{M}$ is homeomorphic to $(\overline{M} \setminus U)/\partial U$, which is is homeomorphic to $\mathbb{S}^3/T$ (where $T$ is a tubular neighborhood of the knot $K$), which is homeomorphic to $\mathbb{S}^3/K$.
Comment: One can fill $\overline{M} \setminus U$ with a solid torus (both have 2-torus as a boundary) in different ways. One of these ways gives you ${\mathbb S}^3$, others give different compact 3-manifolds. Most of these manifolds carry a hyperbolic structure. This is called the hyperbolic Dehn surgery.
