I cleaned up the proof so that it no longer uses the somewhat unprofessional "coset multiplication" argument. I'm happy with it now, so I'll post it as an answer in case it is of interest to anyone.
Recall that the goal is to prove the following:
If $G$ is nilpotent and $N \lhd G$, then $G/N$ is nilpotent.
First, we prove a simple lemma:
Lemma: If $\theta : G \to H$ is an epimorphism, then $\theta(Z(G)) \leq Z(H)$.
Proof: Let $z \in Z(G)$ and $h \in H$. Then $h = \theta(g)$ for some $g \in G$, and $\theta(z)h = \theta(zg) = \theta(gz) = h\theta(z)$. Therefore, $\theta(z) \in Z(H)$.
Now we prove the following proposition:
Proposition: If $G$ is nilpotent and $\phi : G \to H$ is an epimorphism, then $H$ is nilpotent. (Note: Applying this proposition to the canonical epimorphism $G \to G/N$ will yield the desired result.)
Proof: Since $G$ is nilpotent, we have $N_i \lhd G$ and $1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$, with $N_i / N_{i-1} \leq Z(G / N_{i-1})$.
Define $M_i = \phi(N_i)$. We claim that $1 = M_0 \leq M_1 \leq \cdots \leq M_r = H$ is a central series for $H$.
First, we observe that $M_0 = \phi(N_0) = \phi(1) = 1$ and $M_r = \phi(N_r) = \phi(G) = H$. Also, $M_i \lhd H$ by the correspondence theorem, since $N_i \lhd G$.
Now define $\theta : G / N_{i-1} \to H / M_{i-1}$ by $\theta(gN_{i-1}) = \phi(g)M_{i-1}$. It is routine to verify that $\theta$ is a well-defined epimorphism. We can therefore apply the lemma to conclude that $\theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$.
Note also that $\theta(N_i / N_{i-1}) = \phi(N_i) / M_{i-1} = M_i / M_{i-1}$.
Since $G$ is nilpotent, we have $N_i / N_{i-1} \leq Z(G/N_{i-1})$. Consequently, $M_i / M_{i-1} = \theta(N_i / N_{i-1}) \leq \theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$, as required.
You can do this directly, given your definition of nilpotent.
Let $U$ be a proper subgroup of $G$. Then there exists a largest $j$ such that $Z_j\subseteq U$. Since $U$ is proper, $j\lt r$. Then $Z_{j+1}\nsubseteq U$. Since $Z_{j+1}/Z_j\leq Z(G/Z_j)$, it follows that any element in $Z_{j+1}/Z_j$ centralizes $U/Z_j$. Thus, if $x\in Z_{j+1}$ and $u\in U$, then $xux^{-1}\in uZ_j\subseteq U$. Hence $Z_{j+1}$ normalizes $U$, so the normalizer of $U$ contains elements not in $U$ (namely, elements of $Z_{j+1}\setminus U$, which is nonempty since $Z_{j+1}$ is not contained in $U$).
Best Answer
(i) A finite nilpotent group is the direct product of its Sylow-subgroups.
(ii) The quotient of a finite $p$-group by its Frattini subgroup is elementary abelian.
(iii) If the quotient of a finite group by its Frattini subgroup (the set of non-generators) is cyclic then the group itself is cyclic.
(iv) The direct product of two finite cyclic groups of coprime orders is itself cyclic.
From these standard results it is clear that a finite nilpotent group which is not "elementary" must have at least two Sylow subgroups which are non-cyclic, for primes $p,q$ say. Hence we get a quotient $\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_q\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}\times\mathbb{Z}_{pq}$.