When computing relative homology, I know how to find the maps, how to find all the absolute homology groups, and how to show $H_2(X,A)$ is (usually) $0$. My issue is I do not understand how to compute $H_1$ and $H_0$ in this example, but for different reasons than the OP. I do not see where this comes from:
"I use the theorem that says that the succession of $$0\to H_1(\partial M)\to H_1(M)\to H_1(M,\partial M)\to 0$$ is an exact one."
Well, if that is true, $H_1(M,\partial M)$ is immediate, because for any exact $0 \to A \stackrel{f}\to B \to C \to 0$ we have $C\cong B/\text{im}(f)$, and I already know the image there is $2\mathbb{Z}$. I can find $H_0$ similarly.
But I am unfamiliar with this idea of isolating a smaller short exact sequence from a long exact sequence. When can one do this and why?
In the LES for $H_n(M,\partial M)$, the group $H_1(M,\partial M)$ is followed by $\mathbb{Z} \to \mathbb{Z} \to H_0(M,\partial M) \to 0$. Why can we "skip" all this information and "jump" to the last $0$ to obtain the SES above? Or is that not what's going on here?
The easiest way to see it is to use the long exact sequence of reduced homology groups. See for example Definition of $\tilde{H}_n(X,A)$.
Since $\partial M$ is path connected, we have $\tilde H_0(\partial M) = 0$.
If you do not want to use reduced homology groups, you can argue as follows.
$i_* : H_0(\partial M) \to H_0(M)$ is an injection (note 1. $H_0(X)$ is generated by the path components of $X$ 2. The single path component of $\partial M$ is mapped to a unique path component of $M$. This argument does not require $M$ path connected, it may have arbitrarily many path components).
Now look at $H_1(M) \stackrel{j_*}{\to} H_1(M,\partial M) \stackrel{\partial}{\to} H_0(\partial M) \stackrel{i_*}{\to} H_0(M)$. We have $\operatorname{im} \partial = \ker i_* = 0$, thus $\partial$ is the zero map. We conclude that $\operatorname{im} j_* = \ker \partial = H_1(M,\partial M) $, i.e. $j_*$ is a surjection.