I think you just use three subgroups lemma. For subgroups $A,B,C$: $[A,B]=[B,A]$ and $[A,B,C]$ is contained in the normal closure of $[B,C,A][C,A,B]$. One views $K$ as a normal subgroup of the holomorph $KG$ (one should just be careful to take $K$-normal closures; I'll ignore this once or twice to avoid too much notation).
Lemma: Let $K(n)$ be the $K$-normal closure of the subgroup generated by all commutators of weight $n$ with exactly one occurrence of an element of $K$. Set $K[n]$ to be the $K$-normal closure of $[K,G]_{n-1} = [K,G,\dots,G]$. Then $K(n) = K[n]$.
Proof: This is clearly true for $n=1,2,3$; also $K[n] ≤ K(n)$ is clear. Since $K(n)$ is (by definition) generated by the $K$-normal closures of $[K(n-i),G(i)]$ for all $i=1,\dots,n-1$, we need to prove that $P(i): \qquad [K(n-i),G(i)] \leq K[n]$ for $i=1,\dots,n-1$. $P(1)$ is just the statement that $[K(n-1),G(1)] = [K[n-1],G] \leq K[n]$, so we may assume $i \geq 2$ and that (by induction) $P(i-1)$ is true. Then $P(i)$ follows easily: $\begin{align*} [ K(n-i), G(i) ] &= [ G(i), K(n-i) ] \\ &= [ G(i-1), G, K(n-i) ] \\ &≤ [ G, K(n-i), G(i-1) ] [ K(n-i), G(i-1), G ] \\ &= [ K(n-i), G, G(i-1) ] [ K(n-i), G(i-1), G ] \\ &≤ [ K(n-i+1), G(i-1) ] [ K(n-1), G ] \\ &\leq K[n] K[n] = K[n] \end{align*}$
Corollary: In particular, $[G,G,\dots,G,K] ≤ ([K,G,G,\dots,G])^K$, and so if $[K,G]_n = 1$, then $[G,G]_{n-1}$ commutes with $K$, and so if $G$ acts faithfully, $[G,G]_{n-1} = 1$ and $G$ has nilpotency class at most $n-1$.
Probably it is a good idea to understand the case of $K$ an elementary abelian $p$-group, where this is basically just an exercise on upper triangular matrices with 1s on the diagonal.