IRF840 datasheet says Ciss=832pF Input Capacitance @ VDS25V f =1MHz VGS=0 , 1664pF capacitance for two mosfets in AC switch.
I pulled out Qg specifically, because Ciss underestimates it by typically about a factor of 4.
Which... hm, you've already got to be careful here, because there's the original, created by International Rectifier:
https://www.datasheets360.com/pdf/-653557743118012551which says 63nC max Qg and 1300pF typ Ciss. The Fairchild (dated 2002) and the Vishay/Siliconix (2016) datasheets have very similar spec. The ST (2002) version however shows significant improvement with 39nC max Qg and 832pF typ Ciss:
http://www.alphacron.de/download/hardware/IRF840.pdfSo it matters which one you're buying. Not all manufacturers make exact copies of a multi-sourced part!
(FWIW, Vishay is the current owner of the original IR product line, including this part.)
If we use the average capacitance calculation, we get Ceff = (63nC) / (10V) = 6.3nF. Note this is max, so it's not directly comparable to typ Ciss. The Fairchild datasheet gives 42nC typ., or 4.2nF equivalent, 3.5 times the small-signal Ciss figure.
What's going on?
The gate capacitance charges as normal, and is in fact around 1.2nF, and doesn't vary much with voltage by itself. The stinker is the G-D capacitance, which because the drain voltage has a huge swing (up to 500V), this capacitance is hugely amplified (Miller effect). The only reasonable resolution is to look at the charge required to move the gate from 0 to 10V, and divide by that voltage swing to turn it into an average capacitance.
That's the justification for the Qg calculation.

So if the gate looks like 4.2nF (or up to 6.3nF), and it needs to charge in a tiny fraction of 600kHz (1.6us), what resistance is required?
If we say 1/20th of a cycle maximum, or 80ns, in two time constants (t = 2RC), we need a resistance of about 10Ω maximum.
We likewise need to deliver a peak current of (10V) / (10Ω) = 1A, preferably even more to drive it faster. Per transistor, so two in parallel also.
With a ~2:1 current ratio on your drive transformer, that's a demand of over four amperes from the poor ATtiny!
But it cannot deliver anywhere near that current, because the primary circuit has about 280 total ohms. It can deliver about 14mA short circuit.
What could we achieve, then?
This in turn suggests that Fsw should be about as many times slower as the gate drive is. So, if we can't do 4A but we can do 14mA, we should run at a frequency 4/(0.014) times lower, or about 21kHz. And again, lower would be preferable, to save on switching losses.
Or we can add a gate driver IC, but we need to power it at some point. If it's wired directly to the transistors, it needs an isolated supply of 9 to 15V.
However, there is no current limiting resistor on GDT secondary, only turns of copper wire, so its resistance should be quite low - cheap YATO multimeter says: R: 1.3 Ohm 
Yikes! Realize what kind of an assumption this is -- if the secondary had a source resistance of 1.3Ω period, then it must be that, no matter what resistance the primary circuit has, you can draw many times more power from the secondary. You've made a huge power amplifier! That's a very fishy conclusion, so the premise probably was in error.

The equivalent circuit for a transformer may look odd. It's a chunk of wire, right? Well, that chunk of wire is supposed to look like a large AC impedance (that's what the turns around the core does). So it doesn't factor into our circuit. Rather, the primary and secondary sides are connected by the ratio. This looks more obvious if we draw it with no transformer at all -- but to do that, we have to scale all the values from one side, by the ratio, so that they are equivalent to what's on the other side.
Say we scale up the primary side. Assuming 1:2 turns ratio, the primary voltage shows up on the secondary side as double, and the primary current shows up as half. 2 * 1/2 = 1, because energy is conserved of course. This means that the primary resistances appear to go up by a factor of 2*2 or 4. So the MCU and two 100Ω resistors looks like 8V logic (instead of 4V), with ~160Ω equivalent pin resistances, and 400Ω series resistors. A total of almost 1kΩ!
This is why the gate drive will be, needless to say, a bit disappointing.

*Technically, any nonideal transformer can be replaced by an ideal transformer (if needed for 1:1 isolation) and three inductors,
if you don't mind that one of them may be negative. Obviously, negative inductances are hard to come by, and this is just an equivalent circuit. But it's nice to know that equivalent circuits don't stop working just because a number comes out unrealistically.

Now calulated something like average current needed to provide charge for 10Vgs of those IRF840 @ 600kHz and it looks liek this if I do not missed something-I've assumed half of the time charging, second half discharging gate by GDT secondary:
(%i11) F: 600000.0;
(%o11) 600000.0
(%i12) T: 1.0/F;
(%o12) 1.666666666666667e-6
(%i13) T2: T/2;
(%o13) 8.333333333333334e-7
(%i14) t: R*C;
(%o14) 2.1632e-9
(%i15) T2/t;
(%o15) 385.2317554240631
(%i16) Q/T2;
(%o16) 0.019968
Looks like something about ~20mA average current.
Probably by lowering down this switching frequency a few times will help get enougth current for full open/close AC mosfet switch? 
I'm not sure what numbers you used here. Qg(tot) * Fsw does equal supply current for a gate driver IC. The peak current is circa T2/t times higher though.
Tim