Let $P(M,G)$ be a principal $G$-bundle. The action of $G$ on a manifold $F$ is defined from left, the action of $G$ on $P\times F$ is defined as: $$G:\quad(u,f) \to (ug,g^{-1}f)$$
The associated fiber bundle $E=P\times_\rho F$ is defined then as a quotient of $P\times F$ by the relation $\sim$:
$$E = P\times F / \sim$$
where
$$\quad(u,f) \sim (ug,g^{-1}f)$$
Could you please comment on why one cannot simply define $(u,f) \sim (ug,f)$? My natural guess is that this would necessarily lead to the trivial bundle $M\times F$. But probably there are more serious arguments.
The important point in the construction is, that it equips the associated bundle with the transition functions of the principal bundle. To see that recall the following facts:
Since G acts freely and transitively on the Fibers $P_p$, every point in $E=P\times F \Big/\tilde{}$ can be represented as $[s_\alpha(p), x]$ where $s_\alpha \in \Gamma(P\restriction_{U_\alpha})$ is a local section and $x\in F$ is uniquely determined.
A local trivialization $\Phi_\alpha$ on a principal bundle is equivalent to a local section $s_\alpha \in \Gamma(P\restriction_{U_\alpha})$ by: $$ s_\alpha(p)g = \Phi^{-1}(p,g) $$ (Given the trivialization you obtain the section by setting g = e the identity element, and given the section you get the trivialization - This fact is btw very useful for example in seeing that TM isn't trivial by the Hairy Ball theorem).
Loval sections of P are related by the transition functions: $$ s_\alpha = s_\beta g_{\beta \alpha} $$
The local trivializations of $\pi^\rho: E \rightarrow M$ are defined by local sections (<-> local triv of the original principal bundle): $$ (\Phi^\rho_\alpha)^{-1}(p,x) = [s_\alpha(p),x] $$
We thus have: $$ (\Phi_\alpha^\rho)^{-1} (p,x) = [s_\alpha(p),x] = [s_\beta(p)g_{\beta \alpha}, x] = [s_\beta(p), g_{\alpha \beta} x] = (\Phi_\beta^\rho)^{-1}(p, g_{\beta \alpha}x) $$ $$ \Rightarrow \Phi^\rho_\beta \circ (\Phi^\rho_\alpha)^{-1}(p,x) = (p, g_{\beta \alpha} x) $$
So the bundle has the same transition functions as the principal bundle.
It is also instructive to study this construction on the Frame bundle $P= F^GE \rightarrow M$ of which gives you the original vector bundle $E\rightarrow M$