After contemplating the responses to this question (and this one too), I was wondering how to understand why the Möbius band has the center circle as a generator through singular homolgy (seeing as this is the theory Hatcher relies on before reaching example 2.47 in his book Algebraic Topology):
It's supposed to be generated by a singular $1$-simplex that is also a cycle. How can I see there is just one such singular $1$-simplex and that it must be the center circle?
As pointed out in the comments, the generator of $H_1(M)$ for $M$ the Möbius band simply can be taken to be a singular $1$-cycle, since a cycle plus boundary gives a new cycle that's the same in homology.
Your second question in a comment boils down to saying "Why can't the (homology class of the) boundary circle of the Möbius band be taken as the basis element for $H_1(M)$?"
Since $\partial M$ is homeomorphic to a circle, this circle is itself a $1$-cycle, so denote the generator of $H_1(\partial M)$ by $[\partial M]$. Saying that this generates $H_1(M)$ amounts to saying that the inclusion $i: \partial M \to M$ induces an isomorphism on homology, since wanting the boundary circle of the Möbius band to generate $H_1$ means that you want $[i(\partial M)] = i_*[\partial M]$ to generate $H_1$.
Now, consider the composite $\partial M \xrightarrow{i} M \xrightarrow{r} C \to M$, where $r$ is the deformation retraction onto the center circle $C$ and the last map is the inclusion of $C$ into $M$.
If $i_*[\partial M]$ generates $H_1(M)$, then $r_*(i_*[\partial M])$ is twice the generator of $H_1(C)$ since $\partial M$ wraps twice around $C$, and since the inclusion of $C$ into $M$ induces an isomorphism on homology, this composite sends $1 \mapsto 1 \mapsto 2 \mapsto 2$. On the other hand, $M \to C \to M$ is homotopic to the identity map, so this composite should end in $1$, not $2$. This gives a contradiction.
