Thevenin's Theorem is a technique to simplify a complex circuit that defies
ordinary calculation techniques, such that simple ohms law may be applied.
The first step is to identify the cause of complexity, and remove that item from the circuit. If it's removal allows you to calculate voltages and currents using ohms law, you have a Thevenizable circuit. Your task is now to make simple equivalent circuits out of the rest of the surrounding circuitry.
Next re-insert the component you removed from the original circuit into the simplified equivalent Thevenized building blocks, this should, if you chose wisely, result in a simple series circuit that lends it's self to calculation by ohms law.
The technique of Thevenizing the individual portions of the remaining, now separate circuits, is, on a case by case basis, calculate the open terminal voltage, eg. the voltage that would appear if the component that causes the complexity is removed, and write that voltage down. Then short the power supply involved with that group of resistors, that is, replace the power supply with a wire, and calculate the Thevenin resistance, looking back into the circuit, including your wire, where ever it goes. It helps to pretend you're calculating what an Ohm Meter would be reading. Generally the act of doing this results in additional parallel paths, via the "short" you substituted for the power supply. Once you've done that, write down the resistance associated with that portion of the circuit. This resistance, and the associated voltage calculated in the previous step, makes your equivalent circuit, the "Thevenin Black Box". It is a battery, whose voltage is exactly the open terminal voltage calculated earlier, in series with the Thevenized resistance you just obtained.
If there were more than one Black Box, do the rest of them, and finally insert your "temporarily removed component" (in this case a resistor) into the now simplified circuit composed of Black Boxes. Simple Ohms law should now be applicable.
Here's a Lab assignment, construct the "Resistor Bridge" shown above, using resistors of your choice, use values in a range of no more than a ratio of 5 to 1, try to predict the voltage across the center resistor, using Thevenin's Therom, and in doing so remember your open circuit voltage.
Then construct your circuit, and make measurements, both with the resistor connected, and without. Verify that your circuit, and your understanding of it do what you expect them to.
Note: It is possible to wind up with zero volts across the center resistor if that happens the exercise will be rather pointless, so choose a different resistor value in one of the other four.
Also: If you have acquired a 9 volt wallwart don't be surprised if it does not actually produce 9 volts, many produce as much as 12 volts when not subjected to any load, and then when a load is applied their voltage is "pulled" down to become closer to the labeled rating of the wallwart
Incidentally the Black Box power supply for the top picture, is a battery voltage of 3.857 volt, in series with an R-thev of 291.4 ohms, thus the voltage measured at the junction of the 470 ohm, and the Black Box, is 4.294 volts. Other info, the drop across the 470 ohm resistor is 0.7055 volts.