The problem is in this video at 18:00.
If you have a 2-holed 2D torus with a line ($\mathbb{A}^1$) going through one of its holes. How do you deform it to make the line go through both holes.
How to Deform a 2-Torus with a Line Through One Hole
algebraic-topology
Related Solutions
Let $A=\{(x,y,z) \in X : x<\tfrac{1}{2}\}$ and $B=\{(x,y,z) \in X: x>-\tfrac{1}{2}\}$. Then $A$ and $B$ are homeomorphic and each deform retracts onto a surface $\Sigma$ surrounding "half" of the removed subspace; see below. Then $\pi_1(A)\cong \pi_1(B)\cong\pi_1(\Sigma)\cong \mathbb{Z} * \mathbb{Z} *\mathbb{Z}$. We can represent the three generators via paths as in the image below. This choice makes it clear that corresponding generators of $\Sigma_A$ and $\Sigma_B$ are equal when included into $A \cap B$. By van Kampen's Theorem, we have $\pi_1(X)\cong \mathbb{Z}* \mathbb{Z}*\mathbb{Z}$.
Actually that's correct. The fundamental group of the torus is $\mathbb{Z} \times \mathbb{Z}$, and, for example, the subgroup generated by $(1,0) $ and $(0,5)$ is an index 5 subgroup. Indeed it is easy to come up with a covering map $p:T^2 \to T^2$ of arbitrary (finite) degree. As you mention, the group of deck transformations in this picture can be taken to be a cyclic group of rotations.
Best Answer
This would be easiest to describe with a series of pictures but I don't have the time to draw them at the moment. Imagine your 2-holed torus is actually a 1-holed torus with another handle attached to one side along two disks. Now move one of these disks so that it circumambulates the hole of the central torus, moving around the perimeter until it comes back to near where it started. Now both of the torus's holes are penetrated by the line, kind of like two concentric circles which are tangent at a point. If you want you can push one of these circles up a bit so that the line goes through each hole at a different height. I hope this helps!
Update: here is a hand-drawn sequence of pictures which I took a photo of with my cellphone.