This circuit is a VOLTAGE DIVIDER.
Your resistor and the resistance of the voltmeter are connected in SERIES, across the battery, so there is 12V across the series combination. The 12V is divided between the two resistances according to their relative resistances, with the HIGHER resistance having MORE voltage across it.
Digital multimeters usually have an input resistance of 10 megohms (10,000,000 ohms), which is quite high. If you test with a series resistor of 1 kilohm, the TOTAL resistance is 10,001,000 ohms. The fraction (proportion) of the total applied voltage (12V) that is dropped across each resistor is equal to its resistance as a fraction (proportion) of this total resistance.
So the voltage across the 1 kilohm resistor will be 12V x (1000 / 10,001,000)
Which is 0.0012 volts, or 1.2 mV.
And the voltage across the digital voltmeter will be 12V x (10,000,000 / 10,001,000).
Which is 11.9988V. That's 1.2 mV shy of 12V, as you would expect, because the 1k resistor is dropping the 1.2 mV to give a total voltage of 12V.
So you can see that the high input resistance of a digital multimeter on voltage ranges means that it "has little effect" on circuits when you connect it. It doesn't "load down" the circuit much. This is important, obviously; you don't want to disturb the operation of a circuit in the process of measuring it.
Another way of looking at a voltage divider is using
Ohm's law - probably the most important and fundamental law in electronics, although in its simple form, Ohm's law is normally only applied to resistive or "ohmic" circuit elements - and involving CURRENT.
Ohm's law states: I = V / R.
I = current (measured in amps)
V = voltage (measured in volts)
R = resistance (measured in ohms).
When your voltmeter is not connected to your resistor, there is no complete CIRCUIT and no PATH through which CURRENT can flow. When you touch the multimeter probe to the resistor, you complete the circuit, and current flows THROUGH the components as defined by Ohm's law.
The series circuit has a total resistance of 10,001,000 ohms and the total voltage across that resistance is 12V. From Ohm's law, I = 12 / 10,001,000. This works out to 0.00000119988 amps which is 1.19988 microamps (uA unless you can be bothered typing the mu symbol). This is a very small amount of current.
This is the current THROUGH the series circuit consisting of the 1k resistor and the voltmeter. In a SERIES circuit, the current through each part of the circuit is the SAME.
The voltage across each resistance can be calculated by rearranging Ohm's law to state V = I R. In each case, I (current) is the same: 1.19988 uA. But the resistances are different.
For the 1k resistor, V = 1.19988 uA x 1000 ohms
= 1.19988e-6 x 1e3
= 1.19988e-3
= 1.19988 mV.
For the multimeter, V = 1.19988 uA x 10 megohms
= 1.19988e-6 x 10e6
= 11.9988 volts.
These results are the same as before; this calculation introduces the idea of CURRENT FLOW. You can also see that the current is very low, and this is mostly because the multimeter's input resistance is so high. Most DMMs have an input resistance of 10 megohms, sometimes higher. With most circuits, this is high enough for us to assume that connecting the meter will not significantly affect the circuit being measured. Some circuits that operate at very low currents can be disturbed when the DMM is connected. In fact every circuit is affected to some extent, but most are not disturbed.
One analogy for voltage (although this is not the textbook analogy) is distance in one dimension (along a single line). Like distance, voltage is always measured BETWEEN two points. When you think about using a tape measure or a ruler to measure the distance between two parts of your house, you might be surprised if the simple act of measuring the distance actually pulled those points towards each other slightly, but that's what happens when you connect a multimeter to measure voltage in a circuit, because it draws current, however small!
Back in the day, we used multimeters with mechanical indicators (swinging pointer type). These often drew 50 uA from the circuit, at full scale deflection. Connecting the meter would often affect the circuit noticeably.