Your situation is slightly more complicated to put in words because you pull on both masses. This ends up with an action-reaction pair between your hand and block A and a second action-reaection pair between your other hand and block B. Then there's reactions with the spring itself, and it ends up getting a little challenging to book-keep it.
For simplicity, I'd like to offer a different but equivalent problem. Instead of having a spring between two blocks, let's have a spring between block A and the wall, which does not move. Once this scenario is clear to you, it will be easy to extend it to the scenario in your question. But this way I don't have to disambiguate one hand from the other while typing.
So, you pull on the spring. The action reaction pair is always between two objects, and the reaction always has them in the opposite order from the action. So if the action is the hand applies a force on block A (pulling it), the reaction is an equal and opposite force applied by block A to the hand. We can also speak of the action-reaction pair between the block and the spring. The action is the block applying a force to the spring, and the reaction is the spring applying a force to the block.
(As a note: the order doesn't matter. You can assign either one as the action and the other as the reaction. What matters is that they come in pairs.)
Now, you talk of unequal actions and reactions. This cannot occur in Newtonian physics. If you think it's happening, its a good time to slow down and analyze the situation more carefully. You give an important argument for this:
I can pull with any force, while the restoring force is confined to a magnitude of $kx$.
Good. You chose an example to back up why you think the rules don't make sense. We can work with this example.
Rememeber that you're pulling on a block. Let's say you pull on the block with a force $F_{hand}$. As an action reaction pair, the block naturally pulls back on you with a force of $F_{hand}$ (same magnitude, opposite direction). The block is also pulled on by the spring, with a force of $kx$. It too has an action-reaction pair, so it pulls on the spring with a force of $-kx$.
So if we do the force body diagram on the block, we see that it's being pulled on by $F_{hand}$ in one direction and $kx$ in another. These are not action-reaction pairs, so they can be unequal. If they are equal, $\Sigma F=ma$ says that the block will begin to accelerate. Which is exactly what happens. You pull on the block hard enough, and you succeed at moving it outward while stretching the spring.
So what if we make this really tricky, and let you grab the spring directly. Then you apply $F_{hand}$ to the spring. Well, it must apply $F_{hand}$ back at you. But the spring only pulls with a force of $kx$ right?
The trick to this is that you cannot pull on a massless spring with any force greater than $kx$. If you did so, it would extend infinitely fast to the correct length to match your force. Now you'll never see this in real life. All real springs have mass, and don't quite follow Hooke's simple law when stretched really quickly.
This, by the way, is why it hurts so terribly bad to fail to break a board in Karate. If the board breaks, it is not possible for you to have applied a force higher than the breaking strength of the board. But if it doesn't break, you're really only limited by how hard you can punch... which is usually pretty hard because you've been training for it.
And hopefully by this point you're comfortable with the answer to the third question. Take three particles, L (left), C (center), and R (right). By the rules of how forces work, there's an action-reaction pair between L and C (L pulls on C, and C pulls on L). There's an action-reaction pair between C and R (C pulls on R, and R pulls on C). And in both cases, the action and reaction have exactly the same magnitude. They are always balanced. The imbalance you are looking for is that the force of L on C may be unbalanced with the force of R on C, which will cause C to move.
