(Schematic again below.) The cathode current (bias current) from tube C flows through it's 10 ohm cathode resistor to point K, causing a small voltage drop from Ck to K. The similar cathode current from tube D flows through it's 10 ohm resistor to point K, causing a small voltage drop from Dk to K. First, understand that if the bias current from C and from D are exactly equal (meaning bias currents are balanced), the voltage dropped across each 10 ohm resistor will be equal but opposite, so a voltage measurement from Ck to Dk will equal 0V, because they are opposite polarity, therefore cancel.
At point K the two bias currents join, and flow from point K to point T. In doing so, it drops a voltage across the 150 ohm resistor, so that from a measurement of voltage between points T and K we can compute the SUM of the two bias currents. We will adjust the BIAS adj pot until the SUM of the two currents equals 2 times the current we want from each tube.
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In order for the combined bias current to continue flowing from point T to the ground, which is the center wiper of the Balance pot, the current must flow through the 50 ohm BIAS adjustment pot, then divide more or less equally, to flow through both the upper 47 ohm and lower 47 ohm resistors, and through either side of the balance pot to make it to the center wiper and to ground. (This bias current cannot flow through the 470K grid return resistors, so it has to go to ground.) The two currents flowing through these two halves of the DC BAL pot each generate a small voltage drop to ground. This small voltage drop on either side of the BAL pot wiper appears between the 470K grid return resistor and ground for each output tube. Depending upon where the center wiper of the BAL pot is positioned, we can "teeter totter" a small voltage to raise or lower the DC grid voltage of tubes C and D. If tubes C and D were perfectly matched and identical, the bias current of each would be equal with the BAL pot exactly centered so both grids would be at the exact same DC voltage. However, let's say tube C is weaker, so the bias current for that side is smaller, and tube D is stronger so the current on that side is greater. If we move the BAL pot wiper closer to the "D" side, the grid for tube C will be higher, turning the tube on more, and the grid for tube D will be lower, turning that tube down slightly. The adjustment range is not huge here- C and D should be somewhat reasonably matched, or the BAL pot will have to be moved all the way to one side, and the currents still may not balance equally. If C and D are reasonably close, the BAL can achieve equal currents (such that voltage betweek Ck and Dk = 0) somewhere within it's adjustment range.
Setting bias current is a two step operation. Generally you will first power on with the BIAS pot set for maximum resistance (lowest current), and the BAL pot set to the center position. As the amp warms up, you will then first adjust the BIAS pot to bring the SUM of the currents up towards but a little bit short of the final goal, then, you will measure the voltage between Ck and Dk, and adjust the BAL pot to bring it to zero. Then you will measure from T to K and adjust the BIAS pot again closer to the goal, then BAL, back and forth until the tubes are balanced, and at the desired bias current. You will find that one control will have some effect on the other, so you will have to play round robin until everything is set. As the amp continues to warm up, and the line voltage changes, you will see both BIAS and BAL voltage measurement vary and drift around. The key is not perfect aerospace precision, but to keep it all in the ball-park as the amp is operated.
Did I make that clear as mud?