Is there a constant $K \in \mathbb{N}$ such that for every finite solvable group $G$, there exists a nilpotent subgroup $N \leq G$, and a subset $S \subseteq G$ with $S \leq K$, and $\langle N,S\rangle = G$ ?
No. Call the minimal cardinal of a subset $S$ such that $H\cup S$ generates $G$, the (generating) corank of $H$ in $G$. I claim that
There exists solvable (actually metabelian) finite groups for which the smallest corank of a nilpotent subgroup is arbitrary large.
Consider a nonempty finite index set $I$, of cardinal $n$. Consider for $i\in I$ copies $A_i$ of the group with 2 elements, $B_i$ copies of the group of 3 elements. Let $A_i$ act nontrivially on $B_i$ (but trivially on $B_j$ for $j\neq i$). Let $m$ be another integer and consider $$G=G_{m,n}=\left(\prod_i A_i\right)\ltimes \left(\prod_i B_i^m \right)=\prod_i (A_i\ltimes B_i^m)\simeq (C_2\ltimes_{\pm} C_3^m)^n.$$
Let $H$ be a subgroup of $G$.
First assume that $H$ is a 2group. Then conjugating and enlarging $H$, we can assume that $H$ equals the 2Sylow $\prod A_i$. Hence the corank of $H$ is equal to the minimal number of generators of $\prod B_i^m$ as a $(\prod A_i)$module, which is bounded above (actually equal but we don't care) by the minimal number of generators of $C_3^m$ as a $C_2$module, and this number is clearly $m$ (the $C_2$action in this case being scalar, it is useless to reduce the generating rank).
Let now $H$ be arbitrary. Let $J\subset I$ be the set of $i$ such that the projection $H\to A_i$ is onto, and $K$ its complement. Then looking at the abelianization shows that the corank of $H$ is $\ge \#(K)$. Now assume that $H$ is nilpotent. Denote by $N$ the 3torsion in $G$, which is a subgroup (the unique 3Sylow). Then $N\cap H$ is the 3Sylow in $H$; it is contained in $\prod_{i\in K}B_i^m$ (otherwise its projection on $B_j^m$ is nonzero for some $j\in J$ and having in mind that in $H$, elements of order 2 commute with elements of order 3 as in any nilpotent group, we contradict the fact that the projection of $H$ on $A_j$ is nontrivial since the action of the nontrivial element of $A_j$ on $B_j^m$ fixes only zero). Thus after killing $\prod_{i\in K}A_i\ltimes B_i^m$, we are in the case when $H$ is a 2subgroup, and we see that the corank of $H$ is at least equal to $m$, provided that $J$ is not empty. If $J$ is empty, then $n=\#(K)$ and hence the corank of $H$ is at least $n$.
This proves that the corank of any nilpotent subgroup $H$ of $G_{m,n}$ is at least $\min(m,n)$.

$\begingroup$ PS: the 3Sylow $\prod B_i^m$ has corank $n$; the 2Sylow $\prod A_i$ has corank at most $m$ ($=m$ by the above), so it follows that the smallest corank of a nilpotent subgroup in $G_{m,n}$ is exactly $\min(n,m)$. $\endgroup$ – YCor May 6 '15 at 19:58

$\begingroup$ Do you think that it is possible to replace 'nilpotent' by some larger proper subclass of the finite solvable groups, so that the claim will become true? $\endgroup$ – Pablo May 7 '15 at 4:30

$\begingroup$ A candidate could be the class of finite supersolvable groups. $\endgroup$ – Pablo May 7 '15 at 7:04

2$\begingroup$ It lacks of motivation... supersolvable would not work. Replace in the above construction $C_2$ with $C_3$ for $A_i$ and $C_3$ with $C_2^2$ for $B_i$ (with nontrivial $C_3$action). $\endgroup$ – YCor May 7 '15 at 7:35

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