Thanks to Stephen Mitchell's notes on principal bundles, I think I have figured out the relationship. Since any morphism of principal bundles decomposes as a reduction of structure group followed by a pullback, it suffices to consider the case where $\Phi: P \to P'$ is a morphism of principal bundles covering the identity $\mathrm{id}_M$ on the base. As above, $\Phi(u.g)=\Phi(u).\phi(g)$ where $\phi: G \to G'$ is a Lie group homomorphism.
In fact, such a morphism $\Phi: P \to P'$ exists if and only if $P \times_G G' \cong P'$ as $G'$-principal bundles. Indeed, the composite $P \times G \xrightarrow{\Phi \times \mathrm{id}_G} P' \times G \xrightarrow{\rho} P'$ is balanced with respect to $G$, where $\rho$ is the right action on $P'$, so it descends to a map $P \times_G G' \to P'$ out of the balanced product, which is in fact $G'$-equivariant. And since this is a morphism of principal $G'$-bundles covering the identity, it is an isomorphism.
Just like the tensor product for bimodules over noncommutative rings, the balanced product is associative up to natural isomorphism. And by analogy with the classical situation in algebra, the group $G$ is the tensor unit for $(-) \times_G (-)$.
It follows that for any $G'$-rep $V$ (or more generally any $G'$-space $X$) that $$P' \times_{G'} V \cong (P \times_G G') \times_{G'} V \cong P \times_G (G' \times_{G'} V) \cong P \times_G \phi^* V.$$
Since all the isomorphisms here are natural, we have an isomorphism of the associated bundle functors: $P' \times_{G'} (-) \cong P \times_G \phi^* (-)$.