A bogen K10 that came across my path recently met my hacker dark-side; i recall that it had a similar arrangement for g2 power. attached is the schematic for reference.
why is your Vg2 low? if R1 is the upper resistor connected to B+ to g2; R2 is the lower 1/2 of the divider connected from g2 to ground; since current flowing through the Kathode is = to Anode current + g2 current, then g2 is in parallel with R2 (kind of, but for now it's easier to simplify with that view) so, assuming the engineers had picked our tube's parameters beforehand, we would assume that R1/R2 divider network would need to flow at least 10x the current needed by g2 in order to not grossly upset the ratio of the R1/R2 divider network.
you CAN configure the divider network to flow enough current to make just about any bias parameters of your pentode work IF you select values (read: lowER Rs) that flow enough current; the down side with that approach is that you excessively load the PS.
back to your low Vg2, this seems to not be the case if you're truly reading 18V on g2 - if your measurements are indeed correct, then g2 is sucking a lot of current in that bias configuration, or could this just simply be how this stage was engineered, in that this combination of R1/R2 gave the engineers the Vg2 they were looking for.
to summarize: if you have plagiarized an ENTIRE stage and/or you know what your B+, Ia, & Ig2 currents will be by design, then you can make the R1/R2 divider work for you. otherwise, take the high road and use a singe series resistor bypassed to ground + your B+ will love you more.
the RCA Resistance Coupled Amplifier Charts provide a wealth of knowledge - attached are excerpts relevant to the 6AU6.
--DL
EDIT:
IR1 = 278V-18V = 260V so then IR1 = 260V / 150KΩ = 1.733mA
IR2 = 18V / 27KΩ = I of R2 = 667uA
so by KCL then IR1 = Ig2 + IR2. rearranged; Ig2 = IR1 - IR2 = 1.733mA - 667uA = 1.066mA
VRa = 278V-204V = 74V, so then IRa = 74V / 270KΩ = 274uA, yes, that's microamps or read as 0.274mA. you show .65V Vg, so then Ik = .65V / 576Ω = 1.13mA; if current through the cathode resistor = sum of screen current + anode current, then Ik = Ia + Ig2 = 1.066mA + 274uA = 1.34mA which is slightly more than what you show with Vg/Rk (.65V/576Ω). voltmeter loads the grid, measure from cathode to ground, that is very close the -V bias of the control grid. IOW, if you measure 0.65V from cathode (pin 7) to ground then the grid voltage is approximately -0.65V.
would you measure Vdc from pin 7 to ground please? ....and lastly, why the 16 ohm trim pot? the trim pot does seem moot.