"Conservation of momentum is probably a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen."
Sure it does, at least in Newtonian physics where forces act instantaneously (the link in your first sentence, with the comment that the 3rd law is "definitely not universal", deals with electromagnetism, a theory consistent with relativity's requirement that no information can travel faster than light; in relativistic theories the 3rd law no longer holds, though conservation of momentum still does). Force is just the first derivative of momentum $ dp/dt $, meaning the force on a body at any given moment is just the rate of change of the body's momentum at that moment. This implies that if the force $F$ on a body is constant over some time interval $ \Delta t$, then that body's change in momentum must be given by $ \Delta p = F \Delta t $, just by the definitions of force and momentum. If you have two systems 1 and 2 with no external forces acting on them, conservation of momentum implies that over any time interval $ \Delta t $ it must be true that the changes in momentum $ \Delta p_1 $ and $\Delta p_2 $ were equal and opposite, i.e. $ \Delta p_1 = -\Delta p_2 $. So if the forces were constant over the time, it must be true that $ F_1 \Delta t = -F_2 \Delta t $ (where $F_1$ is the force on system 1 and $F_2$ is the force on system 2), so $ F_1 = -F_2 $. Even if the force is non-constant, you can divide the time interval in question into a series of infinitesimal intervals of length $ dt $ such that the force can be taken as constant over each one, and the above argument implies that between any time $ t $ and $ t + dt$, it must be true that $ F_1 (t) = -F_2 (t) $.
By the same token, the force at each moment can be integrated over some time-interval to give the total change in momentum over that interval, and if the forces on two bodies are equal and opposite at each moment then each body must experience an equal and opposite change in momentum over any given time-interval.
You give this example:
"1) Any body can offer a max resistance = k (reaction), any action on it which is < k will get an adequate reaction, but any action which is > k will obviously not get a sufficient reaction. If a ball hits a wall with p > k the wall will crumble."
What does crumbling vs. not crumbling have to do with Newton's third law? Even if the ball punches a hole through the wall, it is still slowed down, and the ball's decrease in forward momentum must have been caused by forces from bits of the wall acting on the ball for the short time interval that it was passing through the wall. At every moment that the ball was passing through the wall, the force that the ball was exerting on each little bit of the wall was matched by an equal and opposite force from that little bit back on the ball.
Edited to add a little more on this, since it's a particular point of contention: Although Alba is unwilling to change his/her mind on this point, for the benefit of other readers I just want to point out that what I said above is definitely the mainstream physics view, for example in my comments to Bert I pointed to the snippets from the solutions manual to a college physics textbook here and here (and some additional snippets which can be seen googling specific phrases) which pose the following question and answer:
A brick hits a window and breaks the glass. Since the brick breaks the
(a) the force on the brick is greater than the force on the glass,
(b) the force on the brick is equal to the force on the glass,
(c) the force on the brick is smaller than the force on the glass,
(d) the force on the brick is in the same direction as the force on
Solution: According to Newton's third law, the answer is (b). Why
isn't answer (c) correct since the brick breaks the glass? It takes a
considerably smaller force to break a glass than to break a brick.
When the brick hits the glass, they exert equal but opposite forces on
each other, say 150 N. It may take only 100 N to break the glass and
1000 N to break the brick. So the glass breaks and the brick remains
Likewise, from the same page, there's this snippet:
There is another misconception concerning the third law. The third law
states that the two forces are equal no matter what. For example, an
egg and a stone collide with each other, the egg breaks and the stone
is intact. Since the egg breaks, we often conclude that the force by
the stone on the egg is greater than the force by the egg on the
stone. This is not so. The forces are always equal. The egg breaks
because it is simply easier to break.
More generally, if you go to google books perform this search for "Newton's third law" along with the word "always", you can find a large number of textbooks saying that the forces between two interacting objects are always equal and opposite, with no exceptions noted for breakage or any other classical scenario. And note that the people here who claim the forces would be unbalanced, like Alba and Bert, never cite any physics expert making this claim, nor do they offer any mathematical derivation of this claim or any experimental evidence to believe it's true--it's a basically crackpot style of argument which rejects scientific evidence/analysis and expert knowledge in favor of personal intuitions.
"2) can an object provide a reaction if the object never gets in contact with it?"
Yes, in Newtonian physics the gravitational force acts instantaneously at a distance, and the gravitational forces between a pair of objects at any given instant are always vectors of equal magnitude but opposite direction. In relativistic theories like classical or quantum electromagnetism, forces no longer act instantaneously but are limited by the speed of light, so the forces on a pair of objects at a given instant need not be equal and opposite--Newton's laws don't really apply in this form, although apparently there is a modified local form of the third law dealing with forces within the field itself, see p. 384 of Relativity Made Relatively Easy on google books here.
On the scenario with the girl pushing the rail and the ball: (Note: I'm leaving this explanation in for now even though you removed the detailed statement of the scenario seen in this version of your question, since you still say "explain how it works when a skater is pushing at the rail or at a basketball" in your question, but you didn't answer my question in comments about whether you still want a detailed analysis or something more conceptual, if you aren't interested in the gory details I can edit.) As I mentioned in comments, "every action has an equal and opposite reaction" is simply a colloquial way of describing the fact that a force from system A onto system B must be matched, at the same instant, by an equal and opposite force from B onto A. It doesn't mean anything more than that, so if you think you see any other implications from the phrase which can't be directly derived from the assumption of equal and opposite forces at each moment, you need to just accept that physicists don't always use words in a common-sense way. In particular, a misunderstanding to ask which force is the "action" and which is the "reaction"--there is no technical distinction between the two, again it's a purely colloquial way of describing the fact that the forces are equal and opposite at any given instant. It's also incorrect to interpret the "action" as an amount of work done rather than a force (work is force*distance), as you did in the first scenario with the girl where you said the "action" was the 210 J of work done--the third law only says that as long as the girl is pushing the ball or rail, the force from her hand on the object she's pushing is matched by an equal and opposite force from the object on her hand (which would move her backwards if she's standing on a frictionless surface, as seems to be assumed in the diagrams where she acquires a backwards momentum equal and opposite to the forward momentum acquired by the object). In my analysis below, I will assume that the forces between the girl and the object being pushed are equal and opposite in this way.
I'd be curious as to what source that diagram actually came from, though, because I think they made a slight error in their numbers for the case of the girl pushing the rail. I would presume they tried to keep things simple by assuming the girl exerted a constant force F on the rail before letting go after some time t. And if they want the total work to be 210 J, then the sum of work done on the rail and work done on the girl should be 210 J. And work is force * distance, so to find the work done on each, we just need to figure out the distance each moved while the force was being exerted. For an object with mass m that starts at rest, if it experiences a constant force F its displacement as a function of time is given by $d(t) = (1/2)(F/m)t^2$, so the work as a function of time is given by W(t)=F*d(t) or $W(t)=(1/2)(F^2/m)t^2$. Since the mass of the girl is 20 kg and the mass of the rail is 1000 kg, and the time and the magnitude of the force should be equal for each, then if we want the sum of work done on each to be 210 J, this gives us the equation $(1/2)(F^2/20)t^2 + (1/2)(F^2/1000)t^2 = 210$. This simplifies to $(51/2000)F^2 t^2 = 210$, and dividing both sides by (51/2000) and then taking the square root of both sides gives $F* t = 90.7485$. Force * time is equal to the impulse, which is just the change in momentum; so the change in momentum should be 90.7485, but they gave the change in momentum as 91.6 instead, I think they made a mistake in their math (or maybe it was just a roundoff error). So, from here on in the analysis I'll use my corrected value for the impulse to the rail rather than the one they give.
Now, if the girl exerted the same constant force F on both objects for two different times $t_1$ and $t_2$, then it would be true that $F * t_1 = 20$ and $F * t_2 = 90.7485$. The first equation can be rearranged as $ F = 20/t_1$, and substituting that value for F into the second equation gives $ 20 (t_2/t_1) = 90.7485$, divide both sides by 20 to get $(t_2/t_1) = 4.5374$. So with the assumption she used the same constant force, she must have pushed the rail for 4.5374 times as long as she pushed the ball--if we imagine F=20 Newtons, then she could have pushed the ball for 1 second and the rail for 4.5374 seconds. Again, Newton's third law just says that during the time period she was pushing each object, the object was pushing backwards on her with an equal and opposite force of 20 Newtons in the other direction.
Are these numbers consistent with the idea that the total work done on both the girl and the object being pushed was 210 J in both cases? Again remembering that work is given by $W(t)=(1/2)(F^2/m)t^2$, let's look at each case:
For the work done on the ball, we have F=20, m=1, and t=1. So the work would be (1/2)(400/1)(1) = 200.
For the work done on the girl by the ball, we have F=20, m=20, and t=1. So the work would be (1/2)(400/20)(1) = 10.
For the work done on the rail, we have F=20, m=1000, and t=4.5374. So the work would be (1/2)(400/1000)(20.588) = 4.12
For the work done on the girl by the rail, we have F=20, m=20, and t=4.5374. So the work would be (1/2)(400/20)(20.588) = 205.88
So, you can see that in each case, the sum of the work done on the girl and the object does indeed work out to 210 J. And you can also see that this analysis was based on assuming that while the girl exerts a 20 N force on the object, the object is exerting a 20 N force back on her, in accord with Newton's third law.
Response to more new edits:
"Can someone please tell me if this is true? According to 3rd law, If I exert a force of k N on 10 different bodies, shouldn't I expect the same reaction of k N from all the 10 different bodies?"
Yes, you should. However, the statement of mine you are asking about, "Saying the third law should imply equal forces in any arbitrary pair of scenarios seems obviously nonsensical", was referring to any arbitrary pair of scenarios, if you restrict the comparison to scenarios where you apply the same force to both objects then in that case the force of the objects on you would be the same. But in the problem with the girl pushing the ball and the rail it wasn't specifically stated that the force she exerted on each was the same--in my analysis, I did assume for simplicity that the instantaneous force was the same, but I also assumed she exerted the force for a longer time period on one object, which was why the change in momentum was different in the two scenarios.
"Consider a man in vacuum or frictionless environment pushing (F = kN) at one object (stone, ball etc.) with one hand or at two objects with two hands in opposite direction, show:
1) that the object(s) exerts an equal reaction (F = kN) and it is a force and not a (change of) momentum
2) that the same force exterted on an object of different mass provokes the same reaction ."
As I asked in a comment, how do you want people to "show" this? Normally it's just assumed as a basic axiom, and the axioms of Newtonian physics are judged based on how predictions derived from these axioms agree with experiment. So do you want experimental evidence or what? You can derive equal and opposite forces from conservation of momentum, if you have two isolated objects that are the only momentum-carriers (no momentum carried by a field as in electromagnetism, for example). But I already did this in the first paragraph of my answer--if you find anything about that derivation to be problematic, please point to the specific step you have doubts about.