This answer is maybe more complete than you want, but hopefully it is beneficial to on-lookers:
Here $U$ is $S^1$ minus the south pole and $V$ is $S^1$ minus the north pole. He is using the Mayer-Vietoris sequence
$H_1(U)\oplus H_1(V)\rightarrow H_1(S^1)\rightarrow H_0(U\cap V)\stackrel{g_*}{\rightarrow} H_0(U)\oplus H_0(V)$
This is an exact sequence, meaning the image of one arrow is exactly the kernel of the next arrow.
The first term $H_1(U)\oplus H_1(V)$ is $0$ because $U$ and $V$ are contractible (so all but $H_0$ vanishes for each of them). Thus the first arrow is the zero map so the second arrow in injective (trivial kernel), and so $H_1(S^1)$ is isomorphic to its image in $H_0(U\cap V)$, which is equal to the kernel of the last map, $g_*$. To understand this kernel, we need to remember what this map does.
In general we have inclusions $i:U\cap V\rightarrow U$ and $j:U\cap V\rightarrow V$, inducing homomorphisms in homology. For $\alpha\in H_0(U\cap V)$, then by definition $g_*(\alpha)=(i_*(\alpha),-j_*(\alpha))$. (The $-$ in the second coordinate is to ensure exactness)
More specifically to our case, $U\cap V$ has two components, so $H_0(U\cap V)\cong \mathbb{Z}\oplus\mathbb{Z}$. This is why he says the elements in this group are of the form $ax+by$. Then $g_*(ax+by)=(i_*(ax+by),-j_*(ax+by))$.
Now, finally, we use the path-connectedness of $U$ to say that $i_*(ax+by)=(a+b)z$ where $z$ generates $H_0(U)\cong\mathbb{Z}$ (draw $S^1$, see what's going on geometrically at this step. Recall that $x$ and $y$ are homology classes of points, and as such $i_*(x)$ and $i_*(y)$, sitting in a path-connected space, are homologous.) It then follows from this last equation that $i_*(ax+by)=0$ iff $a=-b$. By an identical argument, ker$j_* $ has the same condition. Then $ax+by\in$ker$g_*$ iff $a=-b$, and so $H_1(S^1)\cong$ker$g_*\cong\mathbb{Z}$
For your second question, as discussed in the comments up to homotopy you should get a wedge of two circles. Consider the Möbius band as a square with opposite sides identified with a half twist. If we take an interior point away from that square, we can retract onto the boundary "frame". Then if we identify opposite sides we get a "$\Theta$ space," homeomorphic to the Greek letter. Then we can just shrink the line in the middle to make an "$8$"