principal $G$ bundle from principal $H$ bundle given a morphism of Lie groups $\phi:G\rightarrow H$ Let $\phi:G\rightarrow H$ be a morphism of Lie groups. 
Given a principal $G$ bundle, we can associate a principal $H$ bundle by what is called associated fiber bundle for a principal bundle.
Can we do something in other direction in some special case? Suppose $Q\rightarrow M$ is a principal $H$ bundle, can we associate a principal $G$ bundle $P\rightarrow M$ such that if we associate principal $H$ bundle with this, we should get $Q\rightarrow M$.
Can this happen if $\phi$ is some surjective submersion?
 A: In general, no.  That is, there is no way of obtaining a principal $G$-bundle from an $H$-bundle in such a way that the $H$-bundle associated to the $G$-bundle gives back the original $H$-bundle.
For example, set $H = SO(3)$.  There is a non-trivial $SO(3)$ bundle over $S^2$, $SO(3)\rightarrow Q\rightarrow S^2$, which is classified by a map $S^2\rightarrow BSO(3)$.  Since $\pi_2(BSO(3)) = \pi_1(SO(3)) = \mathbb{Z}_2$, there is a unique such non-trivial bundle.
Now, let $G = SU(2)$, $\phi:G\rightarrow H$ the double covering map (which is a surjective submersion.)
Because $SU(2)$ is $2$-connected, $BSU(2)$ is three connected, so any principal $G$-bundle over $S^2$ is trivial.  Thus, whatever procedure you apply to $Q\rightarrow S^2$ results in the trivial bundle $G\times S^2\rightarrow S^2$.
But any bundle associated to a trivial bundle is trivial.  More specifcally, the way of getting back to $H\rightarrow Q\rightarrow S^2$ from $G\rightarrow P\rightarrow S^2$ is that we form the space $(P\times H)/G$ where $G$ acts diagonally on $P$ via the principal bundle structure and it acts on $H$ via multiplication by $\phi(g)$.  Projection onto the first factor defines a map $(P\times H)/G\rightarrow P/G \cong M$, with fiber $H$ (which $H$ acts on by right multiplication).
If $P = G\times S^2$, then $Q = (P\times H)/G = (G\times S^2 \times H)/G = (G\times H)/G\times S^2 \cong H\times S^2$, so $Q$ is trivial.  Thus, there can be no procedure with the property you asked for.
Incidentally, if you insist that both $G$ and $H$ be simply connected, I no longer have any counterexamples, but I would still be surprised if it worked.  (Though it might work in the compact case, owing to the fact that if $G\rightarrow H$ is a surjective morphism of compact simply connected groups, there is a section $H\rightarrow G$)
