How can a resistor affect current AND potential at the same time?
There is no factor that determines if the voltage or the current is reduced.  That whole concept is erroneous.

The simple statement you are looking for is:

A Resistor Defines the Relationship Between the Voltage and the Current

That is, if the current is fixed, then the resistor defines the voltage.  If the voltage is fixed, then the resistor defines the current.

In all three of the Ohm's Law formulae you will have two of the three values as fixed values - values you know, through measurement, or whatever, and the third variable is the one you want to find.  From there it's simple maths.

The LED example, though, throws an extra spanner in the works, since the LED isn't a linear device.  So its influence on the circuit is calculated separately before Ohm's Law is applied.

You have three known values, and you want to calculate a fourth.

The known values you have are: the supply voltage (9V), the LED forward voltage (say, 2.2V as an example), and the current you want to flow through the LED (30mA).

From that you want to calculate the value of the resistor.

So you subtract the LED's forward voltage from the supply voltage, since those are both fixed voltages, and the result will be the amount of voltage that must be dropped across the resistor for the whole to total 9V.  So 9V - 2.2V is 6.8V.  That is a fixed voltage.  The current you want is fixed too - you have decided on 30mA.

So the resistor value is then:
$$
R=\frac{V}{I}
$$
$$
\frac{6.8}{0.03} = 226.\overline{6}  \Omega ≈ 227  \Omega
$$
You will always have two of the three values as fixed values - either because they are set by external factors, like the power supply or battery voltage, or they are a specific value that you require or desire, when it is you who has set that value. The third value is what must be calculated to make both those fixed values hold true.