Intuition behind associated bundles seems natural but I can't see how it's working out "intuitively"! Suppose $L_1\rightarrow P \xrightarrow{\pi} M$ is a $G$-principal bundle with fibers diffeomorphic to $L_1 = G$, and that $\sigma$ is an effective action of $G$ on $L_2$, i.e. if $g.x := \sigma_g(x) = x$ for all $x \in L_2$, then $g = e$ (with $e$ being the identity element of $G$).
I have the following intuition about the associated bundle $L_2 \rightarrow P\times_G L_2 \rightarrow M$. Suppose in above we have a principal $G$-bundle $P$ and for some reason we are interested to replace the fibers that are diffeomorphic to $L_1$, with new fibers that are diffeomorphic to $L_2$. How could we do this?
I was intuitively thinking that if somehow $L_2$ could have been embedded inside $L_1$, then as in the quotient manifold theorem, if $L_2$ represents the quotient space $L_1/G$ for some action $\sigma$, then we might be able to somehow do this: we get rid of the redundant information in each fiber $L_1$ to reduce the already given fiber to something diffeomorphic to $L_2$. But not always do we find ourselves in a situation where $L_2$ could sit inside $L_1$ via an embedding. So, what could we do in these cases?
I think this is why we first consider the bigger fibers $P_x \times L_2$ at each $x \in M$ and build the fiber bundle $P\times L_2$; To somehow fit $L_2$ in a bigger space that is also related to $L_1$. Then
by defining an equivalence relation on this product fibers, we should be able to kill the info of $L_1$ and thus successfully replace our fibers with things that are $L_2$. We can do this by the trivial equivalence relation $(a_1, b) \sim (a_2, b)$ for any $a_1, a_2 \in P_x$ and $b \in L$ but this is very trivial and probably wouldn't give us something interesting. So we move on to something that really engages $L_1$ and $L_2$ together and twists and curves the fibers nontrivially so that we get something cool! We can define an equivalence relation on it by demanding $(a.g, b) \sim (a, g.b)$ for $a \in P, b \in L_2$ and $g \in G$. Here $a.g$ is given by the right transitive action of $G$ on the leaves of the principal bundle. Also $g.b$ is nothing but $\sigma_g(b)$ from the left effective action of $G$ on $L_2$.
The theorem in the associated bundle tells us that after quotienting over this equivalence relation, we obtain the new $G$-fiber bundle $P\times_GL_2$ which has now fibers diffeomorphic to $L_2$.
My question is, although the statement of the theorem is saying that the information of the leaves of $P$ that were diffeomorphic to $L_1$ are gone and it only leaves us with the info of $L_2$, I can't see why is this happening intuitively! The equivalence relation $(a.g, b) \sim (a, g.b)$ seems to treat $L_1$ and $L_2$ very symmetrically. Why is this then that in the end of the day the info of $L_1$ are lost after quotienting but the info of $L_2$ remains? I figure it has something to do with the effectiveness of the action $\sigma$ but for the longest time I haven't figured it out how is this helping?
I wanted to share my intuition about this whole concept of associated bundles. Hopefully it helps to build an insight. I will be very happy if you could help me to find the last piece of this puzzle of mine!

Some more insight: We know that the principal bundle from the beginning is also now the induced bundle of the second and thus induced bundles are also another way of replacing the fibers as well as kind of a left-inverse operation to building associated bundles. In fact the induced principal bundle forgets a huge chunk of information whereas the associated bundle seemingly removes the info if $L_1 = G$ but doesn't actually since the data of $G$ is still stored in the structure group of the associated bundle although its trace us removed from the fibers.

Update: I think the effectiveness of the action is only there so that the associated bundle which is a $G$ fiber bundle to have a unique choice of an induced bundle. I think the effectiveness of $\sigma$ is not needed to delete the info of $G$ at all. Besides (as discussed in Dr. Will Merry's lecture notes) any $G$-fiber bundle that its structure group is not acting effectively on its fibers could be made into one that has an effective action by just adding an equivalence relation. So this does not harm the generality.
 A: "The equivalence relation $(a.g,b)\sim(a,g.b)$ seems to treat $L_1$ and $L_2$ very symmetrically."
Well, there is already an asymmetry hiding here.  Note that the points are elements of $L_2\times P$, not $L_2\times L_1$.  Of course, the $G$ action preserves each leaf $L_1$, so the $G$ action on $L_2\times P$ does preserve each $L_2\times L_1$, but that's not what is being considered via this construction.
Moving past this issue, there is still another asymmetry, arising from the nature of the $G$ action on each of $P$ and $L_2$.
The $G$ action on $P$ is free; the $G$ action on $L_1$ is free and transitive.  To say it another way, given $x,y\in L_1$, there is a unique $g\in G$ with $g.x=y$.
By way of contrast, there is no such guarantee for the $G$ action on $L_2$.  Given $u,v\in L_2$, there may be a $g\in G$ with $g.u = v$, but there doesn't have to be.  And if such a $g$ does exist, it need not be unique.
If you'd like an example of this, consider the usual $S^1$ action on $S^2$ by rotations.  If, say, $u$ and $v$ are both the north pole, then any $g\in S^1$ maps $u$ to $v$, so uniqueness fails terribly.  On the other hand, if $u$ is the north pole and $v$ is any other point, then there is no element $g\in G$ with $g.u=v$, so existence fails.
Finally, there is an asymmetry in the choice of projection.  Sometimes this choice is forced on us: if the action of $G$ on $L_2$ is not free, then the projection $P_x\times_G L_2\rightarrow L_2/G$ need not be a fiber bundle, and $L_2/G$ need not be a manifold, etc.  But if the $G$ action on $L_2$ is free (and you may need to assume more - this is certainly enough if $G$ is compact), then one has two projections, and the choice of the two also introduces an asymmetry.
A: According to the approach given in this answer I can give you the following intuition.
You have a space $M$ that you use to parametrize or to label different copies of a space $L_2$, obtaining this way a bundle that we will denote by $E\to M$. Every fibre $E_x$ is a copy of $L_2$.
You can describe the space $L_2$ in several ways, in fact, you have the set $Diff(L_2)$ of all the diffeomorphism. But maybe they are too many descriptions for you, because some of them may "distort" the space in ways you don't like (think, for example, in $L_2=\mathbb R^n$: you usually are only interested in some special diffeomorphisms, those belonging to $GL(n)\subset Diff(\mathbb R^n)$, the linear maps).
This "reduction" of the space of diffeomorphisms is your $G$. If the group $G$ is not directly contained in $Diff(L_2)$ you require the existence of an effective action $\sigma$ to be able to interpret $G$ as a subset of $Diff(L_2)$, but this is only a technical requirement. Stick to the idea that $G$ is an "interesting" subgroup of $Diff(L_2)$.
Now, what about the principal bundle $P$?
In the same way that every fibre $E_x$ is a copy of $L_2$, every fibre $P_x$ is a copy of the group $G$ of different diffeomorphisms of $L_2$ you are interested in. They are, indeed, not a group, because there is no a privileged identity point. They are what it is called a $G$-torsor.
With this in mind is easy to understand the equivalence relation
$$
(a,b) \sim (a\cdot g, g^{-1}\cdot b),
$$
which is another way to write what you put in your question.
For example, if you think of $L_2$ as $\mathbb{R}^n$ and $G$ as $GL(n)$ then $a$ is a basis and $b$ are coordinates of a vector. It is clear that a vector itself is the equivalence class of this relation: when I use a matrix $g$ to change the basis $a$ to $a\cdot g$, the components with respect to the basis get multiplied by $g^{-1}$. Maybe this link helps you to understand this last point.
