Third law of motion is so confusing According to Newton's third law, whenever objects A and B interact with each other, they exert equal and opposite forces upon each other. I have always struggled with how to apply this law to problems and real life.

Suppose I get a pull from my friend with some force, then I get pulled forward by  a large distance as compared to the distance covered by my friend. Isn't this a case where Newton's third law of motion fails?  Or, does this happen because of the difference in our masses?

My second question is about the tension in the rope in Atwood's Machine (two unequal masses connected by a rope on either side of a friction-less pulley). http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html I have solved various pulley mass problems but I have not thought about applying Newton's Third Law of Motion to it.

Is tension the counterforce to weight? Consider a segment of rope from which a  mass is suspended in Atwood's Machine.  That segment will experience a force $m1g$ from the weight. People say this weight is the reason for the counterforce in that segment, and that this force will travel all the way through the rope to the other segment resulting in a uniform tension.  If this is correct then I can see the uniform tension around the rope as an application of newton's third law but the fact  that this resulted from the weight negates the third law and also this will mean that m1=m2 where m1 and m2 are the masses at each segment
  ...can somebody explain this to me

 A: The motion of you and your friend depends on all the forces which are acting on you you and your masses.  
When you pull your friend forward although the forces between you and your friend are equal in magnitude what about the force that you bother experience due to the Earth?  
Assuming that you had equal masses if net force on your friend is greater than the net force on you then your friend will undergo a greater acceleration.  
Mass will a play a part and so when you start walking towards the front of a moving train the change is speed is not noticeable although it possibly would be if you were in a small boat whose mass is much less than that of the train and thus closer to your mass.  

The Atwood machine with one pulley can be thought of being equivalent to the following example.
 
Where the two masses at the ends of the rope are called $1$ and $3$ and the rope $2$ but you could extend it to being a whole string of as many masses as you wish.  
The masses at the ends of the rope are subjected to gravitational attractive forces and the Newton third law force are labelled $F_{12}$ etc where $F_{12}$ is the force on mass $1$ due to mass $2$, the rope.  
Taking towards the right as positive and applying Newton's second law gives.
$$m_3g-F_{32} = m_3a \;\;\; F_{23}-F_{21} = m_2a  \;\;\;  F_{12}-m_1g = m_1a $$
where is the acceleration of all the masses as the rope is assumed to be inextensible.  
It is often the case that the rope is assumed to be massless which means that $F_{23}-F_{21} = 0 \Rightarrow F_{23} = F_{21}$ and these forces are called the tension in the rope. 
A: In the case of unequal masses attached to a massless rope supported by a frictionless pulley, the tension in the rope is the same everywhere within the rope. Using m1 to represent the smaller mass and m2 to represent the larger mass, and a to represent the rate of acceleration, then
tension = m1 (g + a) = m2 (g - a)
where
a = (m2-m1) g / (m1+m2)
The same formulas apply for m1 >= m2, corresponding to a <= 0.
A: When you pull your friend both experience difference forces, you travel different distances because of several other factors (friction, mass, etc.)
Yes tension is created as a reaction to the weight but only when the mass attached to the string is  not accelerating, in accelerating system the tension is the counter force to the force exerted by the particle on the string (not always equal to the weight).
A: If you and your friend are interacting, when he pulls you, he will feel that you apply to him the same force as he applies to you. If both of you are in vacuum and no further forces are present, the change in momentum will be equal for both of you. If further forces like friction are present, then the total forces acting on each of you might differ. But the parts which come due to you two interacting with each other will still be equal.
In the rope example there are two things playing a role and they should be kept apart:


*

*Each segment of the rope acts on any of the adjacent segments with the same force as the adjacent segment acts on it. Here we are looking at forces acting on different objects interacting with each other. This is Newton's 3rd law.

*The weight and the rope (and thus each segment of the rope) are in a force equilibrium, so the total force on each segment exercised by both adjacent segments totals to zero. Here we are looking at the forces on one object exercised by all other objects it interacts with.


While the first point is always true, the second one doesn't have to be true, say, if the rope and the weight are in a free fall accelerating towards the earth, rather than in an equilibrium situation. Obviously, there will be no tension force in case that the second point is not given.
A: When you are being pulled by your friend, the action is your friend pulling and the reaction is you moving forwards. That's the Third law. Now you move a bit ahead after he stops : because 


*

*Probably the force was increased momentarily before he stopped.

*He kept pulling even after stopping.

*Quoting First Law of motion, an object in motion tends to remain in motion unless an external force acts on it. So you moved an extra distance until the external force of friction made you finally stop.


For simplicity, I consider tension as the opposite of compression. It is like friction. There is a limiting value of tension exceeding which the rope snaps. Tension is indeed the counter force of weight. It acts only when weight is present and that too in the direction of the string, opposite to the weight. Just as a reaction force. As per Third Law, the weight when attached to string must exert a force on the string (action) the string initiated a pulling force in the opposite direction to keep it attached (reaction) which is tension. So there.
It is indeed uniformly distributed in Atwoods machine. Only a massless string would have that. Even if 2kg and 5kg were attached to both ends the tension would have been uniformly distributed. How else would you explain the resultant acceleration if tension were localised? That's why the string moves because of residual force of 30N in my case. Tension balanced 2kg for 20N and 5kg for 50N and the rest caused acceleration. Here too the residual force 30N exerted on the string (action) caused a net acceleration (reaction). I hope this theory helps.
