First differential in cellular homology How do I compute the first differential of cellular homology in general for a CW complex $X$, without using for example that the first differential is zero if $X$ is path-connected and has one zero cell? (Question: doesn't path connectedness follow from having exactly one zero cell?)
By the definition in my lecture, fixing a one cell and a zero cell, we have to consider the map
$$
  S^0 \rightarrow X_0 \rightarrow X_0 / X_{-1} \cong \bigvee_{\text{$0$-cells}} S^0 \rightarrow S^0 \,.
$$
The first map is the attaching map of the one cell, and the last map is the projection corresponding to the fixed zero cell in the wedge sum.
The isomorphism to the wedge sum comes in higher dimensions from identifications $D^k / S^{k-1} \cong S^{k-1}$. But in our case we would have $D^0 / S^{-1} = \mathrm{pt} \neq S^0$. Then it isn't clear how the map $\mathrm{pt} \rightarrow S^0$ would look like, so I cannot determine the degree of the total map $S^0 \rightarrow S^0$.
I don't see how this definition is working out for the first differential or what I can do instead to compute it directly, for example in the case of a $1$-sphere with two $0$-cells and two $1$-cells.

First, we have $S^{-1} = \emptyset$, and by convention (in this context) $X / \emptyset \cong X_+$, i.e., $X$ with a disjoint point added, for any space $X$.  Hence, we have $D^0 / S^{-1} \cong S^0$, which is consistent with the same fact in higher dimensions.
Second, there is an easy way to read off the differential of $1$-cell in (cellular) homology.  In terms of the attaching map $\varphi: S^0 \to X^0$ of a $1$-cell (edge) $e$, write $S^0 \cong \{1, -1\}$.  Then, $$de = \varphi(1) - \varphi(-1).$$ In other words, the differential of a $1$-cell is just the signed sum of its two boundary $0$-cells.
