Understanding this simple circuit

A more systematic approach would be to find the transfer function of the system.

Let's say that the input voltage is given as \$V_{in}\$. Your question now is what happens to \$V_{out}\$ if \$V_{in}\$ goes from 0 to 2V?

To do that, find the transfer function, that is:

$$\frac{V_{out}}{V_{in}} = \frac{R_1||\frac{1}{j\omega C_1}}{R_2 + R_1||\frac{1}{j\omega C_1}}$$

As you said, it's a (complex) voltage divider, expanding the parallel term gives $$R_1||\frac{1}{j\omega C_1} = \frac{R_1 \frac{1}{j\omega C_1}}{R_1 + \frac{1}{j\omega C_1}} = \frac{R_1}{j\omega R_1C_1 +1}$$

$$\frac{V_{out}}{V_{in}} = \frac{R_1||\frac{1}{j\omega C_1}}{R_2 + R_1||\frac{1}{j\omega C_1}}=\frac{R_1}{j\omega R_1C_1 +1} \div( R_2+\frac{R_1}{j\omega R_1C_1 +1}) = \frac{R_1}{j\omega R_1R_2C_1 + R_2 + R_1}$$

This formula is a first order lag element. The general formula of this element is

$$G(j\omega) = \frac{K}{j\omega T + 1}$$

As you can see, the numerator is a polynomial of order 0 and the denominator is a first order polynomial. The coefficients and their values are not that important. That's why this is a first order lag element. The important thing is that many systems have a transfer function like your system. They are all behaving in a similar way, only varying due to values of T, K, etc. They all have a step response of the shape that yours has.

The transfer function is a way to express the behaviour of the system in an abstract mathematical way. It doesn't matter if it is a mechanical or electrical system. If I told you that the flow of heat through an object has the same transfer function, you now know quite a bit about what that means in terms of its behaviour without necessarily knowing anything about thermodynamics. You don't even have to know what heat actually is, because you have an abstract description of its behaviour.

Knowing that is helpful because the step responses for all the basic elements are known. It looks indeed the way you posted it.

The point here is that for more complex circuits, it is often not trivial to figure out what each element doe and how it affects the overall circuit. Analysing the circuit as a circuit with complex elements yields the transfer function that can be used to determine the step response (like in your case), the impulse response or generally speaking the response to any input signal.

Mostly correct.

As pointed out by Joris Groosman you need to look at the thevenin equivalent for this circuit. It looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

I often would tell students to redraw their schematics to help them understand. For you, it may help looking at the initial circuit like this:

^{simulate this circuit}

For me at least, this makes it more obvious that not only will the capacitor be seeing half the voltage, but it will also be seeing half the resistance when looking at the voltage source.

The rest remains the same, when the power supply it turned off the capacitor would be discharging through the lower 1MΩ resistor, following the dotted blue line shown here:

You are correct.

at t=0-, Vout=0, and there is no current, because there no voltage supply. At t=0+ (juuuuuust after you turn the power on), the capacitor acts like a short, so all of the current passes into the capacitor as it starts charging. As the voltage level in the capacitor builds up, a potential across the R1 is formed, and you get a current through the resistor at I=V/R1. Eventually, once the capacitor is fully charges, no current passes to the capacitor and it behaves as an open circuit, so you can ignore it and calculate Vout as if it is a simple voltage divider.