Suppose that $G$ is a group and $N$ is a normal subgroup of $G$. How do we find out what $G/N$ is isomorphic to? For example, $C_m$ is a normal subgroup of $D_{2n}$ generated by a rotation of angle $2\pi/m$. We can partiotion the group $D_{2n}$ using the normal subgrouop $C_m$. We want to find a surjective homomorphism $\phi: G\to G'$ such that $\ker(\phi) = C_m$? What surjective homomorphism $\phi: G\to G'$ should we have so that $D_{2n}/C_m$ is isomorphic to $G'$?
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Define the natural homomorphism $\gamma: G \to G/N$ by $g \mapsto gN.$ Then $N$ is the kernel of $\gamma$ and we have $$G/N\cong \operatorname{Im}(\gamma).$$
See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."
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Usually, the first isomorphism theorem is easily satisfied by the projection map $\pi:G\to G/N$ defined by $$ \pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $\pi(G)$ and get that $$G/N = G/\ker(\pi)\cong\pi(G) $$
I assume $m|n$, so $n=md$ and $C_m\cong\langle r^d\rangle$. In this case, $|D_{2n}/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^{-1}][j]$. The correct guess is that $$D_{2n}/C_m\cong D_{2d}=\langle R,J\mid R^d=J^2=1,JR=R^{-1}J\rangle.$$
Now, define a homomorphism $\phi:D_{2n}\to D_{2d}$ by $\phi(r)=R$ and $\phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_{2n}/\ker\phi\cong D_{2d}.$$ Finally, check that $\ker\phi=\langle r^d\rangle$.