In general, voltage division applies to the voltage across a resistor in series with one or more other resistors.
For example, assume that we have \$N\$ series connected resistors and, further, that the voltage across the series combination is \$V_S\$.
Then, the voltage across the \$n^{th}\$ resistor is given by the general voltage division formula:
$$V_{R_n} = V_S \frac{R_n}{R_1 + R_2 +\; ... \; + R_{N-1} + R_N} = $$
In the diagram you provide, it happens to be the case that
$$V_{out} = V_{R_2}$$
And, by the general voltage division formula applied to two series connected resistors
$$V_{R_2} = V_{in} \frac{R_2}{R_1 + R_2}$$
Thus,
$$V_{out} = V_{in} \frac{R_2}{R_1 + R_2}$$
Note that we could have chosen \$V_{out}\$ to be the voltage across \$R_1\$ instead though this isn't often the case.
In most cases, the output voltage is referenced to ground and, so, the output voltage is taken across the resistor with one terminal grounded.
In summary, voltage division is more general than the two resistor voltage divider circuit you've provided. The principle of voltage division and the general formula is a valuable 'tool' in one's circuit solving 'toolkit'.
If you have a circuit with a 50 V battery and a single 50 ohm resistor there would be a current of 1A flowing through that resistance in accordance with Ohm's Law (V=IR). Now add in a 20 ohm resistor in parallel with this 50 ohm resistor. The current passing through the original 50 ohm resistor will be the same due to Ohm's Law (V=IR): The voltage drop across the resistor is the same (50V) and the resistance is the same (50 ohms) so you would have the same current. However, now the electrons can also flow through the other resistor so the overall current is larger. If you now zoom out and look at the overall circuit you have the same voltage as before but a bigger total current so the total (equivalent) resistance would have to be smaller (since from V=IR for V to be the same if I gets bigger R must get smaller). And this total resistance is given by that equation.
Best Answer
What do you use the push button for? I doubt you need such high current and can easily do with much less, especially when you use it as input for a logic circuit. Although @rawbrawb's 360 ohm will probably work perfectly well, it does draw a lot of current from the power supply. When using a 9V battery, the cell will be drained very quickly.
If the push button indeed is only used for a logic level to drive an opamp or digital chip, a resistor of 10kΩ would already suffice. You didn't share the circuit diagram, that might help in deciding on a good value.