Let $D_{2n} = \langle r,s ~|~ r^n = s^2 = 1, ~ rs = sr^{-1} \rangle$ be the usual presentation of the dihedral group of order $2n$ and let $k$ be a positive integer dividing $n$. Prove that $D_{2n}/\langle r^k \rangle \simeq D_{2k}$.
With respect to link_1, is that solution more complicated than needed? Why not simply note that $|D_{2n}/\langle r^k \rangle | = 2k$, define $\varphi(s^a r^b) = s^a r^b \langle r^k \rangle$, verifying that it is a homomorphism, and then simply note that it is surjective because $s^a r^b \langle r^k \rangle = \varphi (s^a r^b)$, as the author does in the last line? Since we are dealing with a map between finite sets of the same cardinality, it immediately follows $\varphi$ is also injective.
Now with respect to link_2, if they are following Dummit and Foote as I am, and it appears they are, I don't see how the hint would be helpful. According to what I have learned so far in DF, showing that the two generators satisfy the same relations would not suffice to show they are isomorphic (unless I am overlooking something). So, my question is, am I overlooking something in DF, and if not what is the theorem that would guarantee that there would be an isomorpism based on the fact that their generators satisfy the same relations? In short, what would the statement of this theorem be?
EDIT
Here I will show that $\varphi : D_{2k} \rightarrow D_{2n}/\langle r^k \rangle$ as defined above is well-defined. Suppose that $s^a r^b = s^c r^d$ , where these are taken to be arbitrary elements in $D_{2k}$. Now, if either of $a$ or $c$ is zero and not the other, then we have a contradiction, since $s \neq r^z$ for any integer $z$; hence, either $a=c=0$ or $a=c=1$. In either case, the equation reduces to $r^b = r^d$, implying that $b = k \ell + d$. Thus, $\varphi (r^b) = \varphi r^{k \ell + d} = r^{k \ell + d} \langle r^k \rangle = r^d \langle r^k \rangle = \varphi (r^d)$, thereby proving that $\varphi$ is well-defined.
Note that in your way we don't know to which group we're mapping $D_{2n}$. We only know that it's a group of order $2k$, which isn't enough. Hence we need to prove that the group is generated by two elements, namely $\phi(r)$ and $\phi(s)$ and also we need to check that they satisfy the relation $\phi(r)\phi(s) = \phi(s)\phi(r)^{-1}$. This uniquely determines $D_{2k}$
Regarding your second question I this is because in some cases two generators and one relation on them isn't enough to uniquely determine the group. So there are more than one possibility, so we can't guarantee the isomorphism of the groups. But when it comes to the relation $\langle r,s | r^n = s^2 = 1; rs=sr^{-1}\rangle$ it uniquely determines a group.