It is always best to draw a diagram to convince yourself of things in a case like this.
This is intended to represent a steady state situation: nobody is moving / winning. As you can see, there are two horizontal forces on A: the floor (pushing with 100N) and the rope (pulling with 100N). There will be two vertical forces (gravity pulling down on center of mass, and ground pulling up) to balance the torques - I did not show them because they are not relevant to the answer.
Now I drew a dotted line between A and B. Consider this a curtain. A cannot see whether the rope is attached to B (an opponent) or a wall. A can measure the tension in the rope by looking (for instance) at the speed at which a wave travels along the rope - or by including a spring gage.
Now ask yourself this question: if A feels a tension of 100N in the rope (this is the definition of the force on A), and can confirm (by looking at the gage) that the tension is 100 N, but he cannot see whether the rope is attached to a ring or to an opponent, then how can the tension be 200N? If I pull on a gage with a force of 100N, it will read 100N - it cannot read anything else (in a static situation, and where the gage is massless, ... )
I think I understand the source of your confusion based on the earlier q/a that you referenced - so let me draw another diagram:
In this diagram, I have move the point of attachment of the rope with which A pulls B away from B's hands, to his waist. Similarly, the rope with which B pulls on A is moved to A's waist.
What happens? Now there are two distinct points where A experiences a force of 100 N: one, his hands (where he is pulling on the rope attached to B's waits); and another where the rope that B is pulling on is tied around his waist.
The results is that there are two ropes with a tension of 100N each, that together result in a force of 200N on A (two ropes) offset by a force of 200N from the floor, etc.
This is NOT the same thing as the first diagram, where the point on which B's rope is attached is the hands of A - there is only a single line connecting A and B with a tension of 100 N in that case.
As was pointed out in comments, you can put a spring gauge in series with your rope to measure the tension in it; and now the difference between "a single person pulling on a rope attached to a ring at the wall (taken to be the dotted line) and two people pulling across a curtain (so they cannot see what they are doing) is that in one case, a single spring (with spring constant $k$) expands by a length $l$, while in the second case you find a spring that's twice as long, with constant $k/2$), expanding by $2l$.
These are all different ways to look at the same thing.
In part the confusion is because assumptions are made at the start of an analysis and then changed during the analysis.
The simplest system to analyse is an ideal one where the masses are point masses, the string is massless and inextensible, the pulley is massless, there is no friction and the final (steady) state of the system is being considered.
With these assumptions:
For the string the magnitude of the force it exerts on mass $m_1$ at one end, $T_1$, is always exactly equal to the magnitude of the force it exerts on mass $m_2$, $T_2$, at the other end, so $T_1 = T_2 = T$.
Because the string is inextensible the magnitude of the acceleration of one end of the string and hence mass $m_1$ is equal to the magnitude of the acceleration of the string at its other end and hence mass $m_2$, so $a_1=a_2 = a$.
An ideal massless and inextensible string transmits forces between masses and changes the direction of the forces.
Assume that $m_2 > m_1$.
The equations of motion for the two masses can be obtained using Newton's second law:
$m_2 a = m_2g - T$ and $m_1a = T-m_1g$.
With the assumptions that have been made there can be no discussion of how the system reached a state where the acceleration was $a$ from a different set of initial conditions.
If the string had some mass, $m_s$, but was still inextensible then the equations of motion are:
$m_1 a = mg - T_1, m_2a = T_2-m_2g$ and $m_sa = T_1-T_2$.
Note here that as $m_s \rightarrow 0 $ then $T_1 \rightarrow T_2$.
Now consider a real system with the string having a mass, $m_s$ and also capable of being stretched with a "spring" constant $k$.
To make the analysis easier, initially the string has its centre of mass directly over the axle of the pulley, masses $m_1$ and $m_2$ are at rest being held up by upward forces $m_1g$ and $m_2g$ and the string is taut.
The ends of the string exert no forces on the masses.
Remove the upward forces $m_1g$ and $m_2g$ simultaneously and apply Newton's second law.
$m_2 a_2 = m_2 g \Rightarrow a_2 =g, \quad m_1 a_1 = m_1 g \Rightarrow a_1 =g, \quad m_s a_s =0 \Rightarrow a_s = 0$.
The reason for these surprising? accelerations is that it takes time for the information/pulse/wave to travel along the string.
Thus one end of the string does not know what is happening to the other end.
That speed depends on, amongst other things, the "spring" constant of the string and the mass per unit length of the string.
As the "spring" constant gets bigger and the mass per unit length gets smaller so the speed of information transfer gets larger.
Linking this with the ideal massless and inextensible string the implication is that for such an ideal string the information travels at an infinite speed.
This would mean that any changes to acceleration etc happen instantaneously.
For a normal string the speed of information transfer is relatively high, so much so that it is unlikely that those initial accelerations would be observed?
So now you have two masses accelerating downwards at $g$ and the centre of mass of the string at rest.
I do not know exactly what happens before the system settles down to some sort of ideal system behaviour but there must be a time when mass $m_1$ actually stops again and then starts moving upwards.
There may well be some oscillations in the string and perhaps even the masses actually oscillate as well as accelerating in a "clockwise" direction.
With friction and other dissipative forces present those initial "oscillations" will be damped down in a short period of time and again I think that it might be difficult to observe them.
A suggestion is that an apparatus is set up, with unequal masses connected by a spring which goes over a pulley, to see if any of the effects that I have described are visible.
Best Answer
I believe you are thinking "backwards". The acceleration actually tells us what is going on. The downward force on the block (weight and/or pull from a different rope) is a different interaction than the rope tension pulling up. There is no reason to expect them to be the same because they are from different sources. The acceleration is what actually happens when the two forces are acting on the same object. The acceleration tells us that the forces cannot be equal at that moment.
Yes, it can if something is done to change the upward tension force to become equal to the downward force. Consider this based on your statement: A rope holds a 1 kg object in equilibrium. The tension in the rope equals the weight of the object. A 500 g object is added to the 1 kg object. The rope stretches, the tension increases, then the system returns to equilibrium. In order for the rope to stretch, the system briefly accelerated downward because the downward force was greater than the tension force at the moment the 500 g was added. After accelerating downward, the tension force changed and began pulling up with a greater force and stopped the downward motion.
On the other hand, imagine the rope is a piece of stretchy plastic or gum. It holds the 1 kg object in equilibrium, but when the 500 g is added, the objects accelerate downward continually because the stretching doesn't increase the tension, in fact, it might decrease the tension, depending on the gum. From the instantaneous acceleration we can calculate what the tension in the gum is, but by observation is can't be equal to the weight.
The tension is NOT a reaction or Newton 3rd Law force of the gravity. It is a separate force.