What you are saying doesn't make any sense. You are saying that something can't experience a force as long as something else doesn't apply an external force on it. That is correct. But this doesn't mean that the object which wants to get pushed can't make other objects push it. That is what Newton's third law is all about. You can push a rock in space and the rock will push you too because you pushed it. Here the rock is a person who is pushing as well. And note, that this 'pushing' isn't a reaction to the original force. This is a different force.

When you push a rock in space you create a new system that includes the rock's mass; assuming the total force remains the same, this only means that you accelerate more slowly. We only feel differences in force if the entities are discrete and independent.

I absolutely don't follow. Your wording is too vague. Look at this way: You agree that by the third law, forces always come in action-reaction pairs, right? Now, consider the same situation where the two people are pushing on each other. Let's consider these two forces: the force exerted by 'A' on 'B' and the force exerted by 'B' on 'A'. Note here, that I'm talking about the deliberate forces that they exert on each other. These do not form the third law's action-reaction pair. It's entirely B's choice that he is pushing. continued

Since these two do not form an action-reaction pair of each other, there must be two other reaction forces for these two. So now I ask, what are the they? Think about it and then reply.

Except that they do form a pair. To determine how much force each provides individually, you have to look at the net force and then set up equations such that $x + y = 100N$ and $x - y = F_n$.

And how do they form a pair? Didn't I mention that it's entirely B's choice that he is pushing? There are four forces in this situation. Two of which comes from each of them which they did deliberately, and the other two are the reaction forces to each of them as a consequence of the third law. And now you've just brought in two redundant equations into this discussion. What are x and y exactly?

x is A's proportion of the force it's applying, and y is B's proportion. There are only two forces, both adding up to 100N and subtracting to the net force of the entire two-body system.

Okay, forget the equations. Your equations come from the fact that there are only two forces which is not the case. According to you it's because they do form a Newton's third law pair. I've given my reason why they don't. Could you now give yours?

When you look at A and B independently, yes, you have $A_f$ and $A_r$, the forward and reverse forces on A, and $B_f$ and $B_r$, the forward and reverse forces on B. But when you look at the system as a whole, you find that $A_r$ is just another name for $B_f$, and $B_r$ is another name for $A_f$, and it reduces to two forces.

People have upvoted your answer unthinkingly. If the upvoters are reading this, would they mind explaining why this answer makes any sense at all?

If B doesn't actively resist A's force then it becomes part of A. Therefore either B is actively resisting A's force by providing its own, or it becomes part of A and increases A's mass. There is no third option where B becomes joined-yet-separate.

Why are you saying that B becomes part of A's mass? When two objects collide, do you say that they become one mass and then apply equations? No. If this is problematic, then imagine that both of them are pushing with their hands on each others chest. So B experiences a force by A's chest due to the third law and another force by A's hands.

B choosing to pull is immaterial. The example is identical if you tie the rope around B's chest.

@zephyr I'll wholeheartedly accept the answer which explains to me properly why my reasoning is incorrect. And the explanation given by this answer doesn't quite do that. In fact, according to the answer, the only forces considered are the ones that are applied deliberately and not the reaction forces. So not only I find this answer not an explanation as to why my reasoning is incorrect but also wrong.

The ones applied deliberately are the same ones that are reacting. That hasn't changed.

If B decides not to push, then A will experience a force by the reaction force caused by him. You are saying that if B decides to push, then for some reason that deliberate push will become the reaction force. You can see this is wrong because if B wants he can apply 150 N instead which isn't equal to the force applied by A and hence these two deliberate forces can't be third law pairs. I know you are trying to defend that the force should be 100 N, but I just don't find your argument correct.

I keep trying to explain why, but you say "no, I think I'll ignore the math". You can't ignore the math. You can't trust intuition, or common sense, or the fuzzy-wuzzies when it comes to physics. You must listen to the math; it is the only thing that has proven itself correct time and time again.

You are just not giving me a direct explanation! I just gave you a reason why they can't be action-reaction pairs and now you have brought maths into this. Your explanation didn't include any maths to begin with.

-1: Alraxite makes reasonable arguments (he confused me), and the responses to his questions are flippant and wrong. If you have two astronauts carrying nearly infinitely heavy backpacks, with ropes, one attached from A to B's chest and the other from B to A's chest, you wouldn't get confused that the pulls of A and B add up to determine how fast they approach each other. So why is it confusing when they are both pulling on the same rope? To be honest, I agree that it is confusing, but I haven't sorted out why. There is some failure of intuition here among the trained, not among the untrained.

I sorted out the intuition failure--- it might be just mine and Alraxite (but I doubt it). When the two astronauts are pulling on the same rope, the momentum flow is constricted, and the forces are half as what they are when you are pulling on different ropes. This is what was counterintuitive to me and also to Alraxite, for similar reasons--- why should the flow of a conserved quantity care about the details of what particular region the conserved quantity is going through? But it does care, because when momentum is flowing through two force-generators, the net flow is not additive in serial,

But it is additive in parallel. There is no contradiction with energy conservation, because the displacement in the direction of the force is additive in serial and not additive in parallel. This is the intuitive puzzle for Alraxite--- serial vs. parallel springs, and it's not completely trivial in my opinion, and his arguments were over this thing, not over the notion of tension.