Connect a high value resistor from pin 3 = op-amp non-inverting input (OA+) to ground. 10 megohm may be OK but 1M better and lower still better still - but lower will load MK1 more.
What is [very probably] happening is that there is no DC bias to the OA+ input and in the absence of signal the very high impedance input is driven by a combination of inset-offset voltage, input bias current and the prior positive input - all stored in or charging or discharging the capacitance at OA+ to ground.
Oscilloscope probes and fingers (when connected to bodies) tend to make such arrangements work better or differently by providing a small amount of "ground reference" current or a degree of sink impedance.
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
There are basically three ways to do this. One way is two independent, isolated power supplies. Say, two batteries or two bench supplies. Connect them in series, + to -. The connection between the supplies becomes 'zero', and then one supply gives you + and the other gives you -. The second way is to use a bipolar power supply. This is a power supply that is specifically designed to output equal and opposite outputs around ground. It is basically a single-output power supply with an additional output that is 'mirrored' over the ground connection. In this case, the supplies do not need to be isolated, but you will need to use complementary parts (NPN vs. PNP or NMOS vs PMOS pass transistors, 7805 vs 7905 regulators, etc). The third method is to use a single supply with double the voltage and then synthesize a virtual ground halfway in between. See http://tangentsoft.net/elec/vgrounds.html for many useful details on how to build virtual grounds.
Edit: It is also possible to generate a negative supply from a positive supply with some sort of switching DC to DC converter. An inverter would be the simplest, using either a standard inverter topology with an inductor or a flying capacitor charge pump. An isolated flyback converter would also work. These can be procured as complete modules from various suppliers. A charge pump may not provide very much current, so watch out if you decide to use one. Switching supplies can also create quite a bit of switching noise, both directly coupled into the output and also radiated from the high current switching components. This may require filtering and careful design to prevent interference.
Also, please don't rotate your voltage sources upside down; it makes the schematic very confusing. I would recommend moving V2 and V3 next to V1, one above and one below the ground wire, and both with the positive side up. Then connect them over to the amplifier with long wires. This makes the required power supply configuration very obvious.