TITLE: Motivation for construction of associated fiber bundle from a principal bundle
QUESTION [7 upvotes]: Given a principal $G$ bundle $P(M,G)$ and a manifold $F$ with an action of $G$ on it from left, we construct a fibre bundle over $M$ with fiber $F$ and call this the associated fiber bundle for $P(M,G)$.
I do not get the motivation behind the construction given in Kobayashi and Nomizu which I will write down below.
Idea is to construct a fibre bundle with fibre $F$ i.e., we need to construct a smooth manifold $E$ and a smooth map $\pi_E:E\rightarrow M$ that gives a fiber bundle with fiber $F$.
Kobayashi's proof goes as follows : 
They consider the product manifold $P\times F$ with an  action of $G$  as $g.(u,\xi)=(ug,g^{-1}\xi)$.  Then they consider the quotient space $(P\times F)/G$ and call this $E$. 
Consider the map projection map $P\times F\rightarrow M$ defined as $(p,\xi)\mapsto \pi(p)$.
This induces a map $\pi_E:E=(P\times F)/G\rightarrow M$. As $P\rightarrow M$ is a principal $G$ bundle given $x\in M$ there exists an open set $U$ containing $x$ and a local trivialization $\pi^{-1}(U)\rightarrow U\times G$. They then give a bijection $\pi_E^{-1}(U)\rightarrow U\times F$  and give a smooth structure on $E$ so that these bijections are difeomorphisms.  Then, they cal $(E,\pi_E,M,P,F)$ the fiber bundle associated to principal $G$ bundle. 
I am trying to understand the motivation for the above construction.
Suppose $F=H$, a Lie group  and the action of $G$ on $H$ is given by a morphism of Lie grroup $\phi:G\rightarrow H$ with $G\times H\rightarrow H$ given by $(g,h)\mapsto \phi(g)^{-1}h$ do get a principal $H$ bundle in above construction?
Edit : I thinnk above content looks like it is asking why do we need the construction of associated fiber bundles. No, what I am asking is, suppose I have a Principal $G$ bundle $P(M,G)$ with an action of $G$ on a manifold $F$ and I want to associate some fiber bundle on $M$ with fiber $F$. Then, what suggests you to  think of above construction. How does it occur naturally? Are there any other properties of Fiber bundle I should have in mind which suggest this way of construction.

REPLY [1 votes]: I'm late, but I think the answer can be given in a simpler fashion.
A principle bundle is in particular a fibre bundle with no choice of what the fibre should be - it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle with fibres $F$ like $P\times F$ and then dividing out the unwanted fibres $G$: $$E:=(P\times F)/G\quad\text{i.e. we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$
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