Why is the tension on both sides of an Atwood machine identical? 
The field forces $F_{g1}$ and $F_{g2}$ push down on Block 1 and Block 2, respectivley, where
$$F_{g1}=m_1g$$$$F_{g2}=m_2g$$
Since the pully system reverses the direction of each force, wouldn't the following be true?
$$T_1 = F_{g2} = m_2g$$$$T_2 = F_{g1} = m_1g$$
And since $m_1 \neq m_2$, wouldn't $T_1 \neq T_2$?
My textbook states that tension is the same throughout the whole string, but I can't wrap my head around why this is so. If $m_1 \neq m_2$, how could the same force $T$ accelerate both of them an equal amount? Wouldn't Block 2 require a greater force?
 A: Tension in a rope is another word for the force the rope exerts.
Forces always occur in equal and opposite pairs. This is true for a rope. There are two ends. The ends pull two objects toward each other with equal force.
Instead of masses hanging from a pulley, suppose two massive people named 1 and 2, had a tug of war. Suppose 1 pulled harder than 2.
So if 1 is pulling harder on the rope than 2, why is the rope pulling equally hard on 1 and 2? Why doesn't the rope pull back with the same force as each person pulls?
That would happen if 1 and 2 were on opposite sides of a wall and pulled on separate ropes tied to the wall. The wall would pull back just hard enough to keep them both still. It would pull back just as hard as each pulled on the rope.
But in a tug of war, both 1 and 2 will move if 1 pulls harder. They will have the same velocity and acceleration in the direction 1 is pulling, and they will stay a constant distance apart.
Suppose 2 wasn't pulling. He was just standing on frictionless ice and letting 1 pull him with force $F_1$. What is the force in the rope? You can see the force is between $F_1$ and $0$.
It isn't $0$ on 2's end. The rope is pulling him, making him accelerate.
It isn't $0$ on 1's end. He is accelerating forward, but the rope is pulling him back. He would accelerate faster if not for the rope.
Likewise, you can see it isn't $F_1$ on 1's end. It isn't enough to hold 1 still.
It isn't $F_1$ on 2's end. 1 is accelerating a total mass of $m_1 + m_2$. The acceleration for both is $a_1 = a_2 = F_1/(m_1+m_2)$. This is smaller than if 1 stood still and pulled 2 toward him with force $F_1$. 2 would accelerate with $a_2 = F_1/m_2$.
So how big is $T_2
$, the force from the rope that accelerates 2?
$$T_2 = m_2a_2 = F_1 \frac{m_2}{m_1+m_2}$$
How big is $T_1$, the force from the rope that holds 1 back?
$$F_1 - T_1 = m_1a_1 = F_1\frac{m_1}{m_1+m_2}$$
or
$$T_1 = F_1 - F_1\frac{m_1}{m_1+m_2} = F_1\frac{m_2}{m_1+m_2} = T_2$$
You can do similar calculations when both 1 and 2 pull.
A: Equality of the two tensions only holds in the special case of a massless wheel in the pulley. If you read forward in your textbook to the section which deals with massive pullies, you will find that indeed these tensions should not generally be the same and it's only in the special case where the wheel is massless that they work out to be the same.
So in the end, I think you're right to be confused about this...the textbook hasn't yet given you all the tools needed to actually justify the assumption that the tensions work out to be equal.
(By the way, the section with massive pullies is likely titled "torque" or something similar. If not, check for "torque" in the index at the back. The massive pullies will be somewhere around there if the text is at all standard).
A: In addition to the assumption of a massless wheel as Myers points out in his earlier answer, the analysis you refer to also assumes the string is massless and perfectly rigid (no stretching/compression).  With this assumption, the tension at both ends of the string is the same.  Specifically, T1 - T2 = ma where T1 and T2 are the tensions at each end of the string, m is the mass of the string, and a is the acceleration of the string.  As m approaches zero, T2 approaches T1 even though the string is accelerating (a is not zero).
Also, when you later consider the mass of the pully- in which case the tensions at the ends of the string are not the same- you will see the implicit assumption that the tensions in the $\textit{string}$ are the same as the forces of the string on the $\textit{pulley}$.  This is true, but not at all obvious as discussed in detail in the following reference. https://www.researchgate.net/publication/318107848_Force_and_torque_of_a_string_on_a_pulley.
You can search this site for detailed discussions of the case considering the mass of the pulley; e.g., search for "Atwood friction".
A: 
how could the same force T accelerate both of them an equal amount?

This part of your question holds the key - specifically the word acceleration.
Lets start with your second question first: Why are the accelerations the same?
This is a result of the constraint that the string is a fixed length.  As long as this is true then for each metre that the larger mass moves the smaller mass moves one metre in the same time.  The distances travelled are the same in the same time so the speeds are the same and likewise the (magnitudes of the) accelerations.
What your question assumes, without explicitly stating, is that $T_1=m_1g$ and $T_2=m_2g$.
There is no reason that this has to be true and if it were true then the forces on each mass would be balanced and they would not accelerate.
The tension will have a magnitude somewhere between the weights of the two masses so for the heavier mass the net force is down and for the lighter mass the net force is up.
A: 
Since the pully system reverses the direction of each force, wouldn't the following be true?
T1=Fg2=m2g
T2=Fg1=m1g

Technically you are correct. But a few changes have to be made to your statement.
The blocks are not at rest and both travel with acceleration $a$. We can assume that $M_1$ goes up and $M_2$ goes down. Since $M_1$ goes up, it's weight increase ($W_{M_1}=M_1(g+a)$). While $M_2$ goes down, its weight decreases ($W_{M_2}=M_2(g-a)$)
$\therefore T_1=W_{M_2}=M_2(g-a)$ and $T_2=W_{M_1}=M_1(g+a)$(modified version of your quoted equations).
Now, $T_1$ supports $W_{M_{1}}$,
$\therefore T_1=W_{M_{1}}=M_1(g+a)=T_2$
A: Its because the pulley in an atwood's machine is an Ideal pulley. It has no mass and its frictionless. This means that the rope is only going to slip over the pulley freely without rotating it at all.  In that case the rope is completely isolated from the pulley and tension should be uniform throughout.
Note that frictionless means that no friction between rope and pulley
Had it been a rolling pulley (one with friction along rim so that rope does not slip), the tension on both ends would be different due to friction. and it would be that difference that help it turn.
Everywhere in mechanics a frictionless pulley implies that its a "slipping" pulley and not a rolling pulley (as we see in our daily life). These frictionless pulleys are kept only to change the direction of pull keeping the tension in the string same.
