How does the force of tension really work? Here's a diagram of the issue I'm having problem. It's based on an experiment we did in class:

Note that the surface is assumed to be frictionless. Also note that I have assigned a positive and negative direction using that arrow and the plus sign. There are two masses, A and B (A on the track, B hanging) which is tied together by a rope  that passes over a pulley.
I have added in the free-body force diagrams for each mass that I was told were correct.
When we did the experiment in class, the blocks were only motionless when someone was holding block A back. But when it was let go (and that is the situation represented by the diagram I displayed) the whole system accelerated in the positive direction (I have assigned that positive and negative direction to make discussing the problem easier).
The question is, "What is the net force in the entire system (both blocks)?" In other words, what is causing the movement?
(I have also been told that the blocks will have the same magnitude of acceleration. Why is this?)
Anyway, I've been told that the net force is the force of gravity on mass B (the hanging one). But I don't really understand why, even after extensive discussion with various people.
Looking back at those force diagrams: I understand that the force of normal and weight of mass A cancel out, so the only remaining force is tension. And I understand that mass B has tension and weight acting on it.
But here's where it gets tricky:
The tension on A is apparently pulling it in the positive direction, while the tension on B is pulling it in the negative direction. Do those opposite tensions cancel one another, causing force of gravity to become the net force?
There must be some sort of cancellation in play, I figure, because I've been told that when all the forces are summed, you just end up with force of gravity propelling the system.
Some questions I've been asking myself about this have been:

*

*How does the relationship between the force of gravity on mass B and the tension in the rope play into this? Isn't the tension caused by that force of gravity? Doesn't that mean that if tensions cancel, the force of gravity's effect is canceled as well?


*Does the pulley affect tensions? For example, we know that there's a positive tension affecting mass A. Is there still a positive tension in existence on the other side of the pulley, or just the negative tension that's acting on B? Might there be some sort of effect whereby two sets of opposite tensions, one set on each side of the pulley, cancel each other out?


*If you pick any and all points on the rope, would there be two opposing tensions at every one of those points?


*Is tension uniform throughout the rope?


*How might differences in mass between object A and object B (which, sorry if the diagram was misleading in the sizes, can have any mass) play into the tension?
Etc.
 A: The forces will cancel the way you imagine when the system is not moving. That is, in a static situation in which the friction on block A is enough to prevent movement. And with no acceleration the forces must add up to zero.
However, if it starts moving, it must be accelerating (or it could not change from zero) and there must be a net force.
Your blocks are physically connected. Usually we assume a mass-less unstretchable rope, but you can more realistically think of it as a spring that is stiff enough that you can ignore it. The masses are connected and must move together, so naturally the blocks accelerate at the same rate though in different directions. The condition for movement is mg for block B is greater than N*coefficient of dynamic friction for block A - after it starts moving.
The pulley is considered frictionless. It acts like a lever that pivots around the center of the wheel and redirects motion between horizontal and vertical. The tensions are just bookkeeping, like the signs in arithmetic. The tensions can still be equal. The main thing is that the force on A from the rope is greater than the frictional force.
You can use this experiment to find the coefficient of dynamic friction for block A. If you slowly add weight to Block B, you can find both the dynamic friction, and the static friction, or stiction. Was that the purpose?
For understanding tension better, try to imagine yourself on the block and holding the rope. You can feel the tension pulling. Imagine two of you, one on each end. Can one be pulling harder than the other?
A: It's alright, tension is pretty subtle! Let me answer your questions a bit out of order.

  
*
  
*If you pick any and all points on the rope, would there be two opposing tensions at every one of those points?
  
*Is tension uniform throughout the rope?
  

You can think about the rope as a lot of tiny masses connected together by springs; this is a cheap approximation for how tension works on the atomic level, where the springs are stretching chemical bonds and the masses are atoms.
In our simple model of tension, every atom is pulled by a spring on the left and a spring on the right, with forces $T_1$ and $T_2$. Then by Newton's second law,
$$T_1 - T_2 = ma$$
where $m$ is the mass of the atom. Since $m$ is very very tiny compared to the other masses in the problem, we must have
$$T_1 \approx T_2.$$
Applying this to every mass, we conclude that each little spring / chemical bond has approximately the same tension, so we can simply talk about "the tension in the rope". This is a good approximation as long as the total mass of the rope is much smaller than the masses of the blocks.

How might differences in mass between object A and object B (which, sorry if the diagram was misleading in the sizes, can have any mass) play into the tension?

Well, now that we've established that there's a uniform tension $T$, we need to figure out what that tension is. The constraint here is that the rope is taut, which means that it can't be scrunching up or stretching out; that translates to the constraint that the accelerations of the two blocks are equal in magnitude. This equation determines the tension.

How does the relationship between the force of gravity on mass B and the tension in the rope play into this? Isn't the tension caused by that force of gravity? Doesn't that mean that if tensions cancel, the force of gravity's effect is canceled as well?
  (I have also been told that the blocks will have the same magnitude of acceleration. Why is this?)

No, the tension isn't equal to the weight of block B, it's whatever is necessary to satisfy the above constraint. For example, suppose that block B was very very heavy, so the acceleration of the whole system will be close to $g$. In this case, the tension is actually quite small compared to the weight of block B, because you only need a little tension to make block A have the same acceleration.
(In fact, as block B gets infinitely heavy, you can show that the tension doesn't go to infinity -- instead, it becomes the weight of block A! It's neat to try to prove this, and see how it works.)

Does the pulley affect tensions? For example, we know that there's a positive tension affecting mass A. Is there still a positive tension in existence on the other side of the pulley, or just the negative tension that's acting on B? Might there be some sort of effect whereby two sets of opposite tensions, one set on each side of the pulley, cancel each other out?

This is a little tricky to word. The tensions of the two tiny springs attached to each atom approximately cancel out, as shown above. But that doesn't mean that the tension is zero -- all of those springs are still stretched. 
Since the pulley is frictionless, it doesn't have any effect except that it 'turns around' the tension. You can show this by considering the three forces on each atom (two springs, one normal force) which gives $T_1 \approx T_2$ as before. 

Anyway, I've been told that the net force is the force of gravity on mass B (the hanging one). But I don't really understand why, even after extensive discussion with various people.

That's not your fault, the question is just worded badly. There are lots of forces involved in this problem acting on different things: gravity on both blocks, normal on one block, and normal from the pulley on the rope. It's not very clear what "the" net force even means.
A: Actually, your understanding seems quite good, and much of what you wrote goes at least in the right direction. In particular you have done well on the crucial part of getting your free-body diagrams right. Let's take this one step at a time, starting with your first question (which happens to be a good question to answer first:
"I have also been told that the blocks will have the same magnitude of acceleration. Why is this?"
This is fundamentally important for this system: The two blocks are connected by a rope that is assumed to be massless and inextensible. The second part (rope being of fixed length) means that both blocks will move at the same speeds and hence identical accelerations: If that was not the case, their relative distance would have to change, by either the rope becoming slack, or the rope stretching. The latter is prevented by our assumption of the rigid rope, the former cannot happen since Block A must be pulled by a force from the rope.
So, at this point we know that both blocks move in unison, in the sense that their linear velocities and accelerations are the same. There is thus a single acceleration which we'll call $a$. Newton's Second Law then tells us that there must be accelerating forces acting on each of the masses. 
For Block A, this means we have
$$m_A\,a=F_{net, A}=F_T$$
since this block has $F_T$ as the only force acting on it in the direction of the motion. For Block B, we get
$$m_B\,a =F_{net, B}=F_g-F_T$$
since the forces acting on Block B are the weight of the block and the tension force.
We can now add the above two equations to obtain
$$(m_A+m_B)a=F_g$$
(notice that the tension force cancels out; this is exactly the cancellation that people told you about), which allows us to find the acceleration of the blocks,
$$a=\frac{F_g}{m_A+m_B}=\frac{m_B}{m_A+m_B}g.$$
In addition, this equation also shows that it is the force of gravitation that is the relevant "net force" accelerating the entire system (Block A and Block B). 
Finally, with the acceleration $a$ known, we can find the tension in the rope from our first equation above,
$$F_T=m_A\,a=\frac{m_A}{m_A+m_B}F_g=\frac{m_A\,m_B}{m_A+m_B}g.$$
We can look at the limiting cases of very small or large rations of the two masses. If we let $m_A\rightarrow\infty$, then $\frac{m_A\,m_B}{m_A+m_B}\rightarrow m_B$, so the tension in the rope is simply the weight of Block B and the acceleration goes to zero, $a\rightarrow0$. If we let $m_B\rightarrow\infty$, then $\frac{m_A\,m_B}{m_A+m_B}\rightarrow m_A$, so the tension in the rope is simply the weight of Block A and the acceleration becomes $a\rightarrow g$.
On your other questions:


*

*The pulley could affect the tension in two cases: (1) If the pulley has mass (or rather an inertial moment, don't worry, you'll learn about this later), then a force would be needed to accelerate the pulley as well. As a result, you would see different tension forces in the rope before and after the pulley. (2) If the pulley is subject to friction, a force would likewise be required to overcome it, and again the tension in the rope would differ before and after the pulley.

*Yes, we can imagine looking at any cross section of the rope and determining the force of tension there. If the rope has no mass, we will find the exact same force at any point of the rope. So, yes, the tension will be uniform in a massless rope. You may be able to see that this will not be true anymore if the rope has mass: In that case forces are required to accelerate the mass beyond the point we are considering. These forces will therefore be different and vary linearly along the rope.

*You can see the effect of different masses on tension in our last equation above.

