How does the current know how much to flow, before having seen the resistor?

Not sure if this is what you're asking, but yes, when the battery is connected, an electric field wave travels from the battery down the wires to the load. Part of the electrical energy is absorbed by the load (depending on Ohm's law), and the rest is reflected off the load and travels back to the battery, some is absorbed by the battery (Ohm's law again) and some reflects off the battery, etc. Eventually the combination of all the bounces reaches the stable steady-state value that you would expect.

We usually don't think of it this way, because in most circuits it happens too quickly to measure. For long transmission lines it is measurable and important, however. No, the current does not "know" what the load is until the wave reaches it. Until that time, it only knows the characteristic impedance or "surge impedance" of the wires themselves. It doesn't yet know if the other end is a short circuit or an open circuit or some impedance in between. Only when the reflected wave returns can it "know" what's at the other end.

See Circuit Reflection Example and Transmission line effects in high-speed logic systems for examples of lattice diagrams and a graph of how the voltage changes in steps over time.

And in case you don't understand it, in your first circuit, the current is equal at every point in the circuit. A circuit is like a loop of pipework, all filled with water. If you cause the water to flow with a pump at one point, the water at every other point in the loop has to flow at the same rate.

The electric field waves I'm talking about are analogous to pressure/sound waves traveling through the water in the pipe. When you move water at one point in the pipe, the water on the other end of the pipes doesn't change instantly; the disturbance has to propagate through the water at the speed of sound until it reaches the other end.

Since the theory has been covered, I'll go with a rough analogy (Hopefully I'm understanding what you are asking properly, it's not so clear)

Anyway, if you imagine a pump (the battery), some pipes filled with water (the wires), and a section where the pipe narrows (the resistor)
The water is always there, but when you start the pump it creates pressure (voltage) and makes the water flow around the circuit (current). The narrowing of the pipe (resistor) restricts the flow (current) to a certain amount and causes pressure drop across it (voltage across resistor, in this case equal to the battery)

With the second circuit (two resistors in parallel) it is reasonably clear that the same amount of current that flows into the top junction must flow out from the bottom junction (see Kirchoff) If the resistors are the same, then they will share the current equally. this can be though of as one large pipe (wire) splitting into two narrower pipes (resistors) and then fusing back into one large pipe again. If they are unequal, then one will take more flow (current) than the other but the total out will always add up to the total in.

You could ask the same question with the water analogy - how does the water "know" how much to flow? Because it's limited by the pipes width and the pumps pressure.

EDIT - It seems the question being asked is a little different than I supposed initially. The trouble is there are a few different answers (as you can see) at different levels of abstraction, e.g. from Ohms law to Maxwell to Quantum physics. At the individual electron level I think you might have a problem due to particle wave duality and double path (see double slit experiment with photon) mentioned by Majenko.
Note that the reason I said above that "the water is always there" is because the electrons themselves do not flow at ~2/3 the speed of light round a circuit, rather the energy from one is propagated to the next (sort of) and so on. A bit like balls bouncing around randomly and into one another, with a average tendency overall to bounce in the direction of applied potential. A simpler way to think of it is like a line of snooker balls - if you hit the white ball into one end, the energy will be "transmitted" through all the balls (they will not actually change position though), and then ball at the other end will break away.
I have a feeling the quantum explanation might go something like: we can only predict the probability that an individual electron will "choose" one path (or be in one particular area) but the process would not be observable directly (i.e. theoretical physics)

Either way I think this an excellent question and needs a good answer (will try and improve this one if time allows), although at the lowest level may be better dealt with on the physics stack.

At first, the current doesn't really know. Assuming a big cartoony switch in the line, when open, it represents a huge impedance. (Capacitive) charge builds up on either side of it; specifically, electrons crowd the negative terminal and the positive terminal lacks the same number of electrons from normal (image charge). Current flow is negligible (fA*), so there is no potential drop across the resistor. Electrons have no net movement or flow because the electrostatic repulsion with their neighbors, including the big bunch at the switch, is equal to the force from the external electric field bias.

When the switch is first closed, the extra electrons near the switch zip to the other contact, filling in the image charge. Now that there isn't a big bunch of bully electrons refusing to move and pushing back, the rest go ballistic ^{_{(hah! not actually, though)}} and start to zip through the circuit.

Those in and near the resistor meet ... resistance (c'mon; I had to). There aren't nearly as many free electrons or sites, so, not unlike the very large impedance presented earlier by the switch, charge builds up on either end as the impatient buggers jostle for a spot in line. It continues to build up until equilibrium is reached: the electrostatic field from the bunch of electrons waiting to get through the resistor is equal to the external electric field bias.

At this point the current knows how much to flow, and won't change ['til you realize that you put in a 1.3-ohm resistor instead of the 1.3-kohm, and it fries and open circuits again].

If the source were totally removed from the system at first, there would be no initial capacitive charge. An instantaneous connection with the source (DPST switch) would lead to an electric field propagating along the wire near c, accelerating and dragging electrons along with it, and leading to the same leaving-the-football-stadium-type crowding at the resistors. In the case with parallel resistors, however, the doors of said stadium may be of different widths, so the equilibrium currents will differ.