General Nonsense
A force is a current, like an electric current. A current is the rate of flow of a conserved quantity from place to place. The force is the current of momentum from one object to another. Like a regular electric current tells you how much charge is going from point to point, a force tells you how momentum goes from point to point.
When you have 20 Coulombs of charge at one point, and 1 second later, you find 20 coulombs at another point attached by a wire, you can say that 20 Amperes of current moved through the wire. It's not 40 coulombs of current, even though one point lost 20 Coulombs, and the other point gained 20 Coulombs. This is what makes currents currents--- you count the flow of the conserved quantity from one place to the other, you don't double count, by saying "oh, 20 Coulombs disappeared here, and 20 appeared there, so it's 40 Coulombs overall". The flow is defined by how much of the conserved quantity crossed an imaginary surface separating the two points, counting once, with one sign for one direction, and the other sign for the other direction. This is why Newton's 3rd law is true--- the force from A to B is opposite to the force from B to A just because it is a current, and the flow in opposite directions is opposite.
The same is true for a hose carrying water from the main to your yard, at a rate of 8 kg/s. In one second, the main loses 8 kg of water, your yard gains 8kg. But the total rate of flow is not 16 kg/s, just 8. The rate of loss of water in the main is equal to the rate of water mass gain in the yard. For water, you don't get confused with double counting.
In the case of 1 dimensional motion, there is a conserved quantity--- the momentum. The momentum associated to a velocity in the positive x direction is positive, and the momentum associated to a velocity in the negative x direction is negative, so unlike Energy (or nonrelativistically, mass), the conserved quantity is signed. It's like charge, not like mass. The rate of momentum flow is in momentums/second, or Newtons. It's annoying that the unit of momentum is not named after anybody, I'll call it "the Galileo". One Galileo is 1 kg times 1 m/s. A Newton is a Galileo per second, it's Galileos of momentum going from one body to another.
When you have two people on two ends of a rope in a tug of war, pulling with equal strength at 10 N, 10 Galileos of momentum are flowing from one to the other every second, through the rope. The tension in the rope is 10N, and the force on opposite sides is opposite sign, and equal to 10N. Just like the hose, if you slice at any point, 8kg/s of water is moving through the hose, when you slice the rope, anyplace you slice it, there is 10N of tension of each end on the other, which is the current of momentum through the rope.
Your thing
Your confusion comes from the issue of reconciling momentum flows with exertion-feeling, that feeling that one is exerting oneself when one is pushing. The exertion-feeling is not directly related to force, rather to energy consumption and production, and without a good model, it is easy to get confused about the forces.
Consider two astronauts in space, pushing on each other. I will model the astronauts as two identical blocks of mass M, and they have a compressed spring attached to a plate. The plate can be attached to the other astronaut, to the other block directly, or you can attach the plate to the other plate. Further, each block can "choose" whether to push--- meaning that there is a removable rigid rod attached to the plate, which prevents it from moving, which you can remove at any time to allow either astronaut's spring to do work.
I will consider four situations, and then the situation should be clear.
- The two plates are each touching the other astronaut--- each one is pushing on the other, and both are pushing.
- The two plates are each touching the other astronaut, and only one is pushing, the other spring is replaced with is a rigid rod.
- The two plates are pushing on each other, and both astronauts are pushing, meaning both springs are free to expand.
- The two plates are pushing on each other, and only one astronaut is pushing, meaning only one of the springs is free to expand.
In all situations, each spring starts out compressed an amount x, and has spring constant k. The total spring constant in the first case comes from adding springs in parallel, the total spring constant is 2k, and the total energy is
$$ kx^2 $$
The two forces from the two springs always add up, the total force at the initial time is $2kx$, and you get double the force, as your intuition suggested for two astronauts pushing on each other's chests.
The work done by the force converts all the energy to kinetic energy, so that the final velocity (which is equal by symmetry) is:
$$ Mv^2 = kx^2 $$
So that
$$ v = \sqrt{k\over M} x$$
That's the velocity the two astronauts have when they shove off each other pushing on each other's chest, with their arms modelled as springs. That's situation 1.
In situation 2, only one of the astronauts pushing, the force is halved, the initial energy is reduced by a factor of 2, and the final velocity is reduced by a factor of $\sqrt{2}$, which comes from the energy law:
$$ Mv^2 = {kx^2\over 2} $$
Next, consider situation 3. Here the two plates are attached to each other. The springs are in series, so that the effective spring constant is halved. But when both springs are compressed, the amount of displacement is doubled, so that the total energy in the springs is
$$ {k\over 2} {(2x)^2\over 2} = kx^2 $$
just as for the first case. The final velocity is the same as in case 1. But the force is completely different! The momentum flowing through one spring is the same as the momentum flowing through the second spring, and the force is kx, not 2kx.
The final velocity, though, is unchanged--- when both are pushing on each other, they end up moving equally fast even though the force is exactly the same as if only one were pushing. The reason is that the displacements are larger by a compensating factor, so that the work done by the forces is greater.
The situation 4 gives the same velocity as when only one was pushing before, but the force is still exactly the same--- the force is kx, whether both springs are pushing, or only one is pushing. This is the counterintuitive thing that confused you--- it depends on whether you are pushing at the same spot or at a different spot. When the momentum flow is through both force sources, the force doesn't add up over the two sources, but the work still adds up.
In case 4, the final velocity is still reduced by $\sqrt{2}$. So while the end velocity doesn't depend on where you push, only whether both people are pushing, the force you generate to do the pushing does depend on where exactly you are pushing.
This is what is leading to the intuition paradox.
When the two people are pushing on each other's chest, the force is doubled. When they are pushing on each other's hands, the force isn't doubled, but the displacement is doubled, so that the work is still doubled. The work is what is felt as the exertion, and to produce the same final velocity (although not the same initial force) you need to exert yourself twice as hard when pushing alone.
But if you have a rope you want to break, and you're not strong enoug, and you have a friend, tie it to a wall and ask him to pull on the same end as you. Don't put him on one end and you on the other.
This is the correct answer, I wrote nonsense previously. The gist of the other answers to the other problem was not wrong, but they didn't consider the issue of pushing through the same link, or pushing through parallel links. It makes a difference in the force.
You will be punching the feather with a really small force. That doesn't mean your arm is punching lightly. Your arm can have a lot of momentum and internal tension due to internal forces that makes you punch hard, but the actual force is defined on the interaction with another object, in this case: the feather, so the force you apply to the feather is pretty small compared with the maximum force your arm would be capable of doing.
Update: In a comment Jim says:
Unless the collision is elastic, in which case the feather will experience the full force and accelerate to an enormous speed
This is incorrect (although it depends on the definition of "enormous"). Here I will show that the "enormous" speed that you can get for the feather is at most twice that of the arm. If you calculate the final speed for an elastic collision (which I'll leave as homework), you will obtain:
$$v_{feather}=\frac{2v_{arm}}{1+\mu}$$
where $\mu=\frac{m_{feather}}{m_{arm}}$
Thus, in the limit of a negligible feather mass, and in an elastic collision, the speed of the feather will be at most twice that of the arm. The speed will be still less if the collision is not elastic.
UPDATE:
For those who find it difficult to get the equation.
From conservation of momentum: $m_{arm}v_{arm.ini}=m_{arm}v_{arm.final}+m_{feather}v_{feather}$
From conservation of energy: $.5m_{arm}v_{arm.ini}^2=.5m_{arm}v_{arm.final}^2+.5m_{feather}v_{feather}^2$
if you eliminate $v_{arm.final}$ you get
$$v_{feather.initial}=\frac{2v_{arm}}{1+\mu}$$
Best Answer
The force that a block exerts on a person is more similar to the force a person exerts on another person than you might initially realize.
Most of the time when a solid surface like a block exerts a normal force, it deforms a little bit, like a spring. The stress that results from the deformation is providing the force, similar (although not totally identical) to the way a person's muscles change their shape to provide a force. The harder the person pushes, the more the solid surface deforms and the higher the force that each of them exerts (at least up to a point where the solid is pushed so hard that it deforms irreversibly, which you're not considering here).
When viewed this way, there's no reason to expect a different answer for the block + person system than the case of the person + person system.
EDIT: Here's another way to say the same thing. The reaction force mentioned in statements of Newton's 3rd law is not a mysterious force that exists in addition to other forces that you might learn about. The reaction force will always come about in a way that relates to forces that can be studied in other contexts: deformation stresses, gravity, air pressure, and so on, depending on the specific system under consideration.