So we need to construct a natural transformation $\phi : K(I)u^\# \Rightarrow u^*(K(J))$, which are functors from $D_J$ to $E_I$. So for every $d \in D_J$ we need a morphism $K(I)(u^\#(d)) \to u^*(K(J)(d))$ in $E_I$.

1. Since $K(I) \dashv H(I)$, this is equivalent to constructing a morphism $u^\#(d) \to H(I)(u^*(K(J)(d)))$ in $E_J$.
2. Now since $H$ is a fibred functor, $H(I)(u^*(e)) \cong u^\#(H(J)(e))$ for any $e$ so by composition with this isomorphism it is sufficient to construct a morphism $u^\#(d) \to u^\#(H(I)(K(I)(d)))$.
3. $u^\#$ is a functor so it is sufficient to construct a morphism $d \to H(I)(K(I)(d))$
4. So we can just use the unit of the adjunction $K(I) \dashv H(I)$.

The whole process is natural since every step is natural.

So we need to construct a natural transformation $\phi : K(I)u^\# \Rightarrow u^*(K(J))$, which are functors from $D_J$ to $E_I$. So for every $d \in D_J$ we need a morphism $K(I)(u^\#(d)) \to u^*(K(J)(d))$ in $E_I$.

1. Since $K(I) \dashv H(I)$, this is equivalent to constructing a morphism $u^\#(d) \to H(I)(u^*(K(J)(d)))$ in $E_J$.
2. Now since $H$ is a fibred functor, $H(I)(u^*(e)) \cong u^\#(H(J)(e))$ for any $e$ so by composition with this isomorphism it is sufficient to construct a morphism $u^\#(d) \to u^\#(H(J)(K(J)(d)))$.
3. $u^\#$ is a functor so it is sufficient to construct a morphism $d \to H(J)(K(J)(d))$
4. So we can just use the unit of the adjunction $K(J) \dashv H(J)$.

The whole process is natural since every step is natural.