Induction is transitive along subgroup chains

For $F\in {\mathrm{Ind}}_H^G({\mathrm{Ind}}_K^{H}W)$, define $\mathrm{\Psi }(F)(g):=F(g)(e)$. If $k\in K$, then since $k\in H$ as well by [F2], $\mathrm{\Psi }(F)(gk)=F(gk)(e)=(k^{-1}\cdot F(g))(e)=F(g)(k)=k^{-1}\cdot F(g)(e)$, so $\mathrm{\Psi }(F)\in {\mathrm{Ind}}_K^{G}W$.

For $f\in {\mathrm{Ind}}_K^{G}W$, define $\mathrm{\Phi }(f)(g)(h):=f(gh)$ for $g\in G$ and $h\in H$. If $k\in K$, then $\mathrm{\Phi }(f)(g)(hk)=f(ghk)=k^{-1}\cdot f(gh)$, so $\mathrm{\Phi }(f)(g)\in {\mathrm{Ind}}_K^{H}W$; and if $h_0\in H$, then $\mathrm{\Phi }(f)(g{h}_0)(h)=f(g{h}_{0}h)=\mathrm{\Phi }(f)(g)(h_{0}h)$, which is the covariance condition for ${\mathrm{Ind}}_H^G({\mathrm{Ind}}_K^{H}W)$. Thus $\mathrm{\Phi }(f)$ lies in that induced module.

For $f\in {\mathrm{Ind}}_K^{G}W$, $\mathrm{\Psi }(\mathrm{\Phi }(f))(g)=\mathrm{\Phi }(f)(g)(e)=f(g)$, so $\mathrm{\Psi }\circ \mathrm{\Phi }=\mathrm{id}$.

For $F\in {\mathrm{Ind}}_H^G({\mathrm{Ind}}_K^{H}W)$ and $h\in H$, one has $\mathrm{\Phi }(\mathrm{\Psi }(F))(g)(h)=\mathrm{\Psi }(F)(gh)=F(gh)(e)=(h^{-1}\cdot F(g))(e)=F(g)(h)$, so $\mathrm{\Phi }(\mathrm{\Psi }(F))=F$ as functions $G\to {\mathrm{Ind}}_K^{H}W$.

Steps 3.1 and 3.2 show that $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse $G$-equivariant $R$-module isomorphisms. Therefore induction is transitive along $K\le H\le G$.