How to test an ignition key switch voltage drop with an Arduino Uno microcontroller?
As Lundin comments, please learn some basic electronics.
Anyway, here is the most simple circuit.



simulate this circuit &ndash; Schematic created using CircuitLab
It is designed to divide the car voltage for a 3.3V Arduino. If you use a 5V Arduino it will work also, but you might lower R1 from 33k to 18k or 22k.
The voltage divider follows this equation:
$$
\frac{R2}{R1 + R2} = \frac{V_{Arduino}}{V_{Ignition}}
$$
Rearranging for R1 leads to:
$$
R1 = R2 * (\frac{V_{Ignition}}{V_{Arduino}} - 1)
$$
You may select any resistor from the E12 (or even E6) series that has the next higher value.
For example, for \$V_{Arduino} = 3.3V\$ you will get \$R1 \approx 32.42k\Omega\$. So you will use a \$33k\Omega\$ resistor.
Each resistor has a rated maximum power dissipation. One form of calculation is \$P = U^2 / R\$. In the circuit R1 takes the most power because it has the highest voltage. Let's see:
$$
P = (V_{Ignition} - V_{Arduino})^2 / R1
$$
$$
P = (14V - 3.3V) / 33k\Omega \approx 3.5mW
$$
$$
P = (14V - 5V) / 18k\Omega = 4.5mW
$$
Any common resistor will take that power.
EDIT:
This is just the start to make the voltages digestible for the Arduino.
To measure the voltage drop, you will need additionally, at least:

A second voltage divider for the voltage of the battery
A bit bigger R1 to enlarge the range to higher voltages as 14V, if you need to consider this
Precautions against voltage spikes
Filters depending on your frequency requirements, analog or in software
Software to measure both inputs and calculate the difference (some Arduino's microcontroller might have a differential mode)