New answer: Here is a more abstract solution: tensor product of modules and balanced products are both (enriched) coends, defined in exactly the same way. So if you pretend all of the topological groups here are noncommutative rings, and the spaces with group action are (bi)modules, then almost everything here is obvious from basic abstract algebra and the extension-restriction of scalars adjunction.
Why are tensor products and balanced products coends? Well a coend can be used to define as a tensor product of functors. Given a covariant functor $F: \mathsf{C} \to \mathsf{D}$ and a contravariant functor $G: \mathsf{C}^{\mathrm{op}} \to \mathsf{D}$ we can take their coend $F \otimes_\mathsf{C} G$ provided $\mathsf{D}$ is monoidal and sufficiently nice.
A ring $R$ can be thought of as a one-object ringoid $R$, and a left $R$-module $V$ is then precisely an additive functor $V: R \to \mathsf{Ab}$. A right $R$-module $W$ is a contravariant additive functor $W: R^{\mathrm{op}} \to \mathsf{Ab}$. Then the coend $V \otimes_R W$ is precisely the usual tensor product of modules.
Likewise a group is a one-object groupoid, and a left $G$-space $X$ is a functor $X: G \to \mathsf{Top}$. Thinking of them as functors, given a left $G$-space $X$ and a right $G$-space $Y$ their coend is the usual balanced product $X \times_G Y$.
Original answer:
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. The converse is easy to obtain by sticking the $G$-equivariant inclusion $P \xrightarrow{\langle \mathrm{id}_P, e \rangle} P \times G'$ in the obvious spot in the above diagram.
Moreover, it looks like (though I have not had a chance to check yet) that in fact this construction is part of an adjunction. This is analogous to the extension and restriction of scalars adjunction for modules.
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: $P' \times_{G'} (-) \cong P \times_G \phi^* (-)$.
