Sluckey has the right approach, or you could copy a Fender blackface bias rectifier ckt.
You say "But inserting any resistor where the diagram shows it burns the resistor immediately.
Shouldn't. You have let's call it 80 volts. a 22K resistor in there (across the + and - ) should pass 80/22,000 = 3.6mils. Power = .0036*.0036 * 22,000 = .287 watts. But you don't especially want a resistor there anyway.
" Adding a pot in series with the output negative voltage does nothing." <<--it won't unless there is a load on the supply. On your breadboard, you are (probably) applying no load. No load = no current flows. No current = no volts will be dropped across a series resistor.
A typical bias circuit in a tube amp draws tiny, miniscule current. What you want to do is to set up a voltage divider; 2 or 3 series resistors, across the filtered output of your bias supply. Let's say 2 resistors. If those resistors are equal value, then the midpoint will be 1/2 your full 80 volts = 40 volts. The trick, usually, is to set up such a voltage divider using THREE resistors, the middle one being a pot such that you can adjust the voltage that appears on the wiper of the pot from let's say -30 volts to -60 volts. This way, you cannot accidentally adjust the bias too low (which would redplate the output tubes) but you CAN apply too much negative bias so as to almost shut down the output tubes. The good setting of the bias pot is roughly in the middle. If your pot is 25K and the resistor above is 25K and the resistor below is 25K, then your pot should cover the middle third of your 80 volts. From 27 to 54 volts. That's pretty close and perfectly usable right there. I like making the one "above" (from the negative filtered output on the bias supply) a bit lower, say 15K. Not necessary. All 3 equal would work fine. And you don't specifically need a 25K pot. Almost any linear pot will do (prefer not 1 MEG or higher)