Prove that if $G$ is finite and $G$ has composition series $1=N_0\lhd...\lhd N_r=G$ and $1=M_0\lhd M_1\lhd M_2=G$ then $r=2$. 
Prove that if $G$ is finite and $G$ has composition series $1=N_0\lhd...\lhd N_r=G$ and $1=M_0\lhd M_1\lhd M_2=G$ then $r=2.$

So if $r=1$ we would have $G$ is simple since $G/1$ is simple, which contradicts that $M_1\lhd G$.
So assume $r>2$ then there must exists some $N_j$ in the first composition series such that $\vert N_j\vert<\vert M_1\vert$ or $\vert N_j\vert>\vert M_1\vert$
Then by second isomorphism theorem we know that $N_j\cap M_1$ is normal in both $M_1$ and $N_j$. Which by correspondence theorem gives me $N_j\cap M_1/M_0\lhd M_1/M_0$. And $N_j\cap M_1/N_{j-1}\lhd N_j/N_{j-1}$
What I need to know is that $N_j\cap M_1$ isn't trivial so that one of these composition factors isn't simple. Which I believe if $N_j\cap M_1/M_0$ is a trivial normal subgroup then either $N_j\cap M_1=M_1$ or $N_j\cap M_1=M_0=1$, but then $N_j\cap M_1/N_{j-1}$ will only be trivial if $N_j\cap M_1=N_{j-1}$ or $N_j$, so this is a contradiction that the $N_i$ were a composition series.

You are almost there. Let’s be a bit more precise in choosing $i$.
(I will assume that you have proper inclusion in your series.)
Note that $1=N_0\cap M_1\unlhd N_1\cap M_1\unlhd N_2\cap M_1\unlhd\cdots \unlhd N_r\cap M_1 = M_1$. Because $M_1$ is simple (since $1=M_0\triangleleft M_1 \triangleleft M_2=G$ is a composition series), in this subnormal series we have that there exists $i$, $0\leq i\lt r$ such that $N_i\cap M_1=\{e\}$, and $N_{i+1}\cap M_1 = M_1$. So $N_i$ is the “last” normal subgroup that has trivial intersection with $N_1$; and $M_1$ is therefore contained in $N_{i+1}$.
Then $\frac{N_{i+1}}{M_1}\triangleleft \frac{M_2}{M_1}=\frac{G}{M_1}$. But since $M_1/M_1$ is simple, either $N_{i+1}=M_1$, or else $N_{i+1}=G$.
I’ll leave it to you to complete the argument that if $N_{i+1}=M_1$ then we must have $i=0$ and $r=2$.
If $N_{i+1}=G$, then $i=r-1$. Now, because $M_1\cap N_j=\{e\}$ for $j=0,\ldots,r-1$, then $M_1N_j$ properly contains $M_1$ for $j=1,\ldots,r-1$. Also, $M_1N_j\unlhd M_1N_{j+1}$, since $N_j\triangleleft N_{j+1}$ and $M_1$ is normal in $G$. So we have
$$M_1=M_1N_0 \lhd M_1N_1\unlhd\cdots\unlhd M_1N_{r-1}\triangleleft M_1N_r=G.$$
But $G/M_1$ is simple. So...

By the way, some of these ideas are behind Schreier’s Refinement Theorem, and the Jordan-Holder theorem; no doubt D&F are noddding in that direction.
