I am adding the answer I received by email from Doctor Tadashi Tokieda, he is the director of studies in mathematics at Trinity Hall, University of Cambridge, some bio here and here, I was following his Topology and Geometry open lectures at Youtube for the AIMS, so I dared to send him an email yesterday (I did not expect an answer, just tried) and received his answer!! (thank you very much Doctor Tokieda, it was a honor!)
He sent me a very nice easy to follow explanation of the concepts, so I am transcribing exclusively the explanation here as a complement (more adapted to layman terms) to the good answer of @AloizioMacedo. The drawings are mine, so I apologize if they are not very accurate. Here it is:
Let us take the example of a disk. Think of the disk as a billiard table and of its boundary circle as the cushion rails. If I understood aright, your picture is: shoot a particle, it slides along a straight line, but when it hits a cushion rail, it reflects, and slides along another straight line... Capture this picture by some sort of gluing diagram.
How about this? Take two copies of the disk, $A$ and $B$. Sew $A$ and $B$ together along their boundary circles, like a pita bread. Puff inside a little, so that the resulting closed surface $S$ looks like a very squashed sphere, with $A$ as the northern hemi-pita and $B$ as the southern hemi-pita. At every point along the equator of $S$, the surface is smooth and its tangent plane is vertical.
Imagine a particle sliding on $S$, and watch it from "above". While it is sliding on $A$, it appears to travel along more or less a straight line on a disk. As the particle approaches the equator, crosses the equator, and gets into $B$, you see its image from above approach the boundary circle of the disk, reflect off the boundary, and continue traveling on the disk more or less along a straight line. We had to say "more or less", because the curved $A$ and $B$ mean that the images of the trajectories seen from above are not exactly straight. But you agree that, as we squash $S$ more or more flat, the trajectories converge to straight lines. In summary, the picture you want on a domain (our example was a disk) can be realized by doubling the domain into a closed surface. The dynamical behavior you want on the domain is the projection of the dynamics on that surface.
There is a whole sub-field of mathematics called "billiards". There is also a well-studied topic of geodesics (trajectories of sliding particles) on Riemannian manifolds. Our discussion above shows that you can reduce the former to the latter, by taking the "squashing" limit.
See: Dynamical billiards at Wikipedia.
Best Answer
The trick is to remember that the entire knot is going to be pushed off to the vertex of the complex. The $1$-skeleton of the gluing is actually those two little "connecting" lines where the knot twists against itself. The $2$-skeleton will be gluing triangles in an "obvious" way, but the edges of the triangles will be glued to those little connecting lines, not to the knot.
Here's a picture I drew: (Disclaimer: I ran out of time so I'm not completely sure that what I labeled "inside" is actually the inside of the tetrahedron.)
Also, if you want to work backwards, the explicit face-pairing is given in the online notes in ch. 1 and ch. 4. I don't remember if it's given in the book version.