I would suggest wiring both ends of the pot to port pins that can be configured not to burn quiescent current while sitting at half-rail (many processors have pins that can be configured to be either digital outputs or analog inputs) and float both ends of the pot while not taking readings. Connect the midpoint of the pot to an ADC input, with a significant-value cap to ground. If you have multiple pots and can't afford to use three processor pins for each one, you could e.g. use external chips to connect and disconnect the ends of the pots. Note that the ends of the various pots should not be tied together; when a pot is disconnected, it should not connect to anything else.
If you do things this way, the ADC input voltage, with its cap to ground, should remain roughly constant when the pot is connected and disconnected. When powering up the cap, take readings repeatedly until either they stabilize or it becomes clear they won't (because they go up and down, rather than asymptotically approaching some value). If there isn't too much leakage in the circuit, you shouldn't have to have the pot powered up very long if it isn't being moved, since the cap should start out at, and remain at, the proper voltage.
What you are missing is that the pot is wired such that it produces a ratio of the input signal with the ratio varying from 0 to 1 accross the full sweep of the pot. This is why the 5 kΩ and 10 kΩ pots resulted in the same full volume.
The pot achieves this by being a resistor divider. It does not work by adding resistance in series with a signal. A resistor divider looks like this:
The output will be R2/(R1+R2) of the output. In the case of a pot, R1 and R2 are one continuous resistor with a mechanical wiper picking off OUT at some point along this resistor. The three pins of the pot are the two ends of this resistor and the wiper tap. Therefore, R2 will vary from 0 at no volume to (R1+R2) at maximum volume. Also, R1+R2 is always fixed, and is the resistance value specified for the pot. In your "5 kΩ" pot, for example, R1+R2 is 5 kΩ, which is the value of the physical resistor that the wiper slides over.
At half volume, for example with the 5 kΩ pot, R1 and R2 are each 2.5 kΩ. OUT is half of whatever signal is applied at IN. Note that since everything is ratiometric, you get the same answer whether the total pot resistance is 5 kΩ or 10 kΩ. This is why the volume levels didn't change.
The total pot resistance does matter in other ways to the driving circuit and whatever is using the signal at OUT. The 5 kΩ pot will require whatever is driving IN to provide twice the current than is necessary with the 10 kΩ pot. You don't know what exactly is driving IN and what its design constraints might have been, so it is best to replace the pot with one of the same value. It seems you got lucky in that whatever is driving IN can cope with the 5 kΩ load, but it could just as well have started clipping, otherwise distorting, or have the frequency balance different.
The crackling and the fact that you got sudden jumps in volume were due to the old pot being worn out. As pots age, dirt and oxidation accumulates on the surface of the resistor where the wiper slides over it. The resistor itself can also get worn down due to mechanical abrasion by the wiper. The wiper sometimes making good contact and sometimes not can sound like crackling, especially when the pot is being turned. Dead and worn out spots on the resistor can cause sudden jumps. These are all common failure modes of pots.
This is one area where construction quality makes a big difference. El-cheapo pots wear out a lot faster and may not be as well sealed against dirt or the materials are more prone to oxidation. If you want long-lived mechanical volume controls, you have to spend the money on good quality pots.
This is also one reason these things are done digitally nowadays. You can get a microcontroller to handle the audio stream digitally for less than the price of a top quality volume control. The digital multiplies inside the micro don't wear out, crackle, or drift over time.
Best Answer
Here's how, if I have understood it right.