I am trying to calculate the degrees of the attaching map of the two cell of the torus. I have the following cell structure:
The $2$-cell is $e_2$ and the $0$-cell (all four corners) is $e_0$.
I called the attaching map of $e_2$ $f_2$ and defined it as
$$ f_2 : \partial e_2 \to X^{(1)}, e^{i\theta} \mapsto e^{i\theta}$$
The $1$-skeleton looks like a figure eight.
This is where I got stuck. To find the degrees of $f_2$ at $e_1^1$ and $e_1^2$ respectively one would use the maps
$$ \partial e_2 \to X^{(1)} \to X^{(1)}/(X^{(1)}\setminus e^1_1)$$
$$ \partial e_2 \to X^{(1)} \to X^{(1)}/(X^{(1)}\setminus e^2_1)$$
The space $X^{(1)}/(X^{(1)}\setminus e^1_1)$ is a copy of $S^1$ because removing a $1$-cell from the figure eight leaves a copy of $S^1$ and taking the quotient of the figure eight by $S^1$ gives a copy of $S^1$.
Therefore the degree of $f_2$ equals $1$ at both $1$-cells.
The problem with this is that for the boundary map I then get
$$d_2 = e_1^1 + e_1^2$$
when I should be getting
$$ d_2 = e_1^1 - e_1^2 -e_1^1 + e_1^2 = 0$$
What am I doing wrong? How can I calculate the degrees of this attaching map?
EDIT: I tried to improve my answer by first clarifying your method, and giving more details on a second method. I hope this helps!
First of all, it is commonplace to use the superscripts for dimension, so one might more typically label the 1-cells $e^1_1$ and $e^1_2$. (This is done in Hatcher, in particular.)
Method 1: Your issue amounts to misinterpreting the attaching maps.
The picture you draw actually indicates immediately that $\partial e^2 = e_1^1+e_2^1-e_1^1-e_2^1=0$. One way to see that is to imagine for a moment you have not yet identified the 1-cells. Then you could say the map was $\partial e^2 = e_1^1+e_2^1-e_3^1-e_4^1$. But in the torus we have $e_1^1=e_3^1$ and $e_2^1=e_4^1$.
Perhaps the key point of confusion is that while the boundary map does amount to two maps $S^1 \to S^1$, the maps are NOT degree 1. After all, maps $S^1 \to S^1$ could take any integer as the degree. In your case, $e^2$ is attached to each 1-cell TWICE, so the degrees could be either 1+1 or 1-1, depending on the orientation.
Method 2: If you prefer a slightly more rigorous method than using pictures, note that for the purposes of homology it is sufficient to describe a map up to homotopy. So, we can describe the map by the induced behavior on fundamental groups.
We have $\partial e^2 = S^1$. The fundamental group of the circle is generated by the (homotopy class of the) loop $ \phi : [0,1] \to S^1, t \mapsto e^{2\pi i t}$. In this example, $X^{(1)}$ is a wedge of two circles, with fundamental group isomorphic to the free group generated by $\alpha,\beta$, which represent the homotopy classes of loops around each of the circle factors.
So we can describe the attaching map $\partial e^2 \to X^{(1)}$ by $[\phi] \mapsto \alpha \beta \alpha^{-1} \beta^{-1}$ in the fundamental groups.
The first homology group $H_1 (X)$ is the abelianization of the fundamental group, so $\alpha \beta \alpha^{-1} \beta^{-1}$ becomes $[\alpha] + [\beta] - [\alpha] - [\beta]=0$.