Show $G=\langle\delta\rangle\ltimes D$ is nilpotent of class $2$.

Note that saying that the group $G$ is nilpotent $G$ is equivalent to saying that the sequence of subgroups defined by $$\gamma_{1}(G) = G, \qquad \gamma_{i+1}(G) = [\gamma_{i}(G), G] \text{, for $i \ge 1$}$$ terminates to $\{1\}$. Here $[H, K] = \langle [h, k] : h \in H, k \in K \rangle$, where $[h, k] = h^{-1} k^{-1} h k$ is the commutator of $h$ and $k$.

Note that the commutator subgroup $\gamma_{2}(G) = [G, G]$ of $G$ is contained in $D$, as $G/D$ is cyclic.

Now consider the commutator $$ [(a_{1}, a_{2}), \delta] = (a_{1}, a_{2})^{-1} (a_{1}, a_{2})^{\delta} = (a_{1}^{-1}, a_{2}^{-1}) (a_{1}, a_{1} a_{2}) = (1, a_{1}). $$ Therefore $$ [(a_{1}, a_{2}), \delta, \delta] = 1. $$ Please let me know if this keeps you going.