This is how I viewed the fender TS with the 6800R ground end lifted AND with if the volume pot is @ "10" and the treble pot is swept down all the way:
the bass pot, the .047 and the 6800 become non-players. high frequencies that see the CR 250pf/1M as low impedance would pass unattenuated (> 600Hz ?) and lower frequencies would go the .1uf route and find divider with R1=250K, and R2=1M. Is this accurate?

Swap the positions of the 250pF cap and the series 100kΩ & 0.1uF cap so that they match
a schematic. Now it should be easier to follow what happens...
If the Treble pot is swept all the way
up, then there is direct connection from the 250pF cap to the Volume pot (as shown in the diagram above). The Treble pot is also a voltage divider between 2 paths, so its position matters. The Bass pot is a rheostat between the ends of a 0.1uF cap and a 0.047uF cap which are
not connected to the 100kΩ slope resistor. Looking carefully at the
layout, you'll see the 2 caps have 0Ω between them at the minimum-Bass setting. So your drawing reflects a state when the Bass pot is set to a minimum, and the cap should be 0.147uF total (both caps in parallel). Both caps and the Bass pot are still in play...
The connection to the tube grid is immaterial, because the Volume pot will just divide the voltage down from the maximum Volume setting; either way, sweeping the Volume pot doesn't change the resistance-to-ground seen by the tone circuit.
Check out
PRR's Reactance Reminder thread for information on how to calculate cap reactance for various frequencies. We'll use that information to replace the caps with their equivalent resistance (equal to their reactance) at 2 representative frequencies: 80Hz for the guitar's low E and 5kHz for harmonics/overtones of the guitar's treble range (it's very likely your speaker will have audible falling response above 6kHz).
80Hz
250pF cap has a reactance of 159/(0.08kHz * 0.00025uF) = ~8MΩ
0.147uF cap has a reactance of 159/(0.08kHz * 0.147uF) = ~292Ω
5kHz
250pF cap has a reactance of 159/(5kHz * 0.00025uF) = ~0Ω
0.147uF cap has a reactance of 159/(5kHz * 0.147uF) = ~0Ω
Read the thread and do the math to verify for yourself. Now replace each cap with a "resistor" for the specific case for each frequency.
At 80Hz, the previous stage's plate sees a path through the 250pF cap of 8MΩ to the top of the 1MΩ volume pot. There is a 2nd parallel path through the 100kΩ in series with ~292Ω, for a total of 1,000,292Ω. The total parallel resistance of the 2 paths will be 8MΩ || 1.000292MΩ = ~890kΩ. This forms a divider with the 1MΩ Volume pot to cut 80Hz to 1MΩ/(1MΩ+890kΩ) = ~53% of the 80Hz output of the previous stage plate. This is 20 log 0.53 = ~-5.5dB.
At 5kHz, the previous stage's plate sees a path through the 250pF cap of 0Ω to the top of the 1MΩ volume pot. The 2nd parallel path through the 100kΩ has substantially 0Ω in series, for a total of 100kΩ. The total parallel resistance of the 2 paths will be 0Ω because the 250pF cap's low reactance shunts the path through the 100kΩ resistor. There is essentially 0dB of loss through the tone network at 5kHz.
~5-6dB of difference between the 2 paths is sensibly close to what the plots showed for Mid pot disconnected from ground, and Treble control at maximum. I had the Tone Stack Calculator set for Bass halfway up, which accounts for the ~1dB less cut.
Now do the same thing for the Treble control at minimum and Bass at minimum. The previous stage plate sees a path through the 250pF cap, then 250kΩ resistance from the Treble pot, then the top of the Volume control. The other parallel path is 100kΩ in series with 0.147uF then the top of the Volume pot.
At 5kHz, both caps are short-circuits, so there is 250kΩ in parallel with 100kΩ (~71kΩ), then the top of the Volume pot for almost zero loss. At 80Hz, there is a path of 8.25MΩ through the 250pF cap in parallel with 0.100,292MΩ through the 0.147uF composite cap. Total resistance looks like ~100kΩ. Previous stage plate output is cut to ~91% its original value, or less than 1dB. So overall frequency repsonse is substantially flat; this again jives with the Tone Stack Calculator result.
You can break down any tone circuit into an equivalent circuit in this manner, though I'd recommend picking extreme settings of the tone controls in some cases to simplify the equivalent circuit you're working with.