Got it. Thanks for taking the time to draw some notes on the schematic, as the instructions are really for a non-tech soldering parts into an amp while I think in terms of a schematic which shows what the circuit is doing.
... If my interpretation is correct then the guitar cable insertion should only disconnect the 68K/1M path to ground. ...
You actually sorted the answer out for yourself!
Signal path is like this:
1st gain stage with 820Ω cathode resistor & 250µF cap, with 0.0022µF coupling cap
1MΩ Volume control -> Wiper to jacks & grid of 2nd gain stage.
2nd gain stage with 2.7kΩ cathode resistor & 0.68µF cap, with 0.022µF coupling cap
1MΩ volume control with 5000pF bright cap -> wiper to
another voltage divider of 470kΩ/470kΩ resistors (with a 500pF bright cap)
3rd gain stage
Cathode follower tone circuit driver
The volume controls are voltage dividers. The jacks on the output of the 1st volume control are an "extra load" which changes how the 1MΩ pot divides voltage.
With nothing plugged in those jacks, the switches on the jacks place the 68kΩ resistors in parallel and connected to ground. The net result looks like 34kΩ to ground.
Let's say you set that first volume control to divide the voltage by half, with 500kΩ from input to wiper and 500kΩ from wiper to ground. The jacks and their resistors are in parallel with the lower half of the pot from wiper to ground, so that 34kΩ is in parallel with the pot's 500kΩ. The total looks like ~32kΩ to ground. Instead of reducing the voltage to 50% of the input voltage, this volume pot setting reduces the voltage to 32kΩ/(500kΩ+32kΩ) = ~6% of the input voltage.
Now you insert a plug into the correct jack with the unused cord, it opens the jack's switch which un-shorts the 1MΩ resistor; the two 68kΩ resistors are still in parallel. The total resistance rises to 1.034MΩ instead of the previous 34kΩ. In parallel with 500kΩ from the pot's wiper to ground, this looks like 337kΩ to ground. With the same volume pot setting as before, output climbs to 337kΩ/(337kΩ+500kΩ) = 40% of the input voltage.