That was a long question :-)
You are not out of your mind and think in the right direction.
In fact you can get twice much amps you need for your fan.
So yes, get one of these modules, and add some extra capacitors on the output (some 10'000 uF) to help it deal with FAN's pulsed current consumption & startup current.
You can even get way bigger fan, cap will help to start it and it will work even if you have just 30-40% of the power it needs. You just need separate switch for DCDC board and FAN - so that you turn DCDC first, and after 1 second - FAN (so that cap is charged).
And finally, direct sun gives much much more juice than 60 Watt lamp. So you really need more powerful fan & more powerful DCDC.
but might 36v from a pair of panels damage the actuator circuitry?
So here's the deal. Lead-acid batteries look electrically like a voltage source/sink with a small series resistance, with the voltage level a function of state of charge. 2V/cell (there are 6 cells in series in a 12V battery) is nominal, and if I remember right, their open circuit voltage is something like 1.9V empty, 2.1V full. That covers 90% of their behavior.
Considering that, the "1W@18V" spec of the solar panel isn't going to be able to "win" against the battery, and the solar panel's voltage will be pulled down to battery voltage, delivering probably 0.055A (=1W/18V) at whatever the battery voltage is.
When a battery gets completely full, however, its series resistance goes up dramatically, and the voltage goes up, until there's enough voltage to start electrolysis of the fluid and you get H2 and O2 generation at the terminals and loss of the electrolyte. A lead-acid battery, depending on the type + manufacturer, has a certain recombination rate of H2 + O2 => electrolyte that it can handle; if you electrolyze at a higher current than that, it leads to permanent electrolyte loss (+hence capacity loss)
So there is a safe current that can be delivered to a lead-acid battery continuously, where its own self discharge due to electrolysis balances the charging current. It depends on the manufacture + construction. I wouldn't feel worried about a C/10 or C/20 rate of charge (where C = the current needed to discharge a battery in 1 hour). Garage door batteries are probably > 1Ah capacity so you should be safe with 55mA charging current.
HOWEVER -- I would probably put a (zener diode and resistor in series) in parallel with each battery, the zener diode being about 14V and resistor being maybe 10 ohms or so, so that it keeps the battery terminals from getting charged too far.
Also: if you can, wire each solar panel to each battery (and keep the diodes), rather than the pair of panels in series wired to the batteries in series -- i.e. try to connect the center taps. By doing so, you'll charge each battery independently. Otherwise, what can ruin battery life is if the battery voltages diverge -- the one with the higher voltage will tend to get overcharged, while the other one will tend to get overdischarged and not completely charged.
Best Answer
I do not know what "blue cell which looks like cracked glass" you are talking about, but I have seen cells on sale rated up to 173 W/m2 nominal power - these where best quality, selected cells. The ratings are however usually given as open circuit voltage times short circuit current, and even with the best cells, the most power you are ever going to get even with the cell facing directly the sun, is 70% of that, which is about 120 W/m2. But that nominal power you are only going to get at full sunlight with the cell facing exactly in the direction of the sun. Yesterday I was making tests with some small cell around 14:00, and when the cell was in horizontal position, I was able to get about 67% of the maximum power it had facing the sun. However when the cell was in shade, the output was only about 6% of the maximum power.
If you are interested in getting 12 V & 40 A only at clear sky, if you buy the best quality cell of rated nominal power 173 W/m2 (real maximum power 120 W/m2), the rough calculation would be:
$$12 \mbox{ V} \times 40 \mbox{ A} = 480 \mbox{ W}$$
$$\frac{480 \mbox{ W}}{120 \mbox{ W}/\mbox{ m}^2 \times 67\%} \approx 6 \mbox{ m}^2$$
That might fit on a van if it is a moderately large one, but if you want it to provide that power also when there are clouds, or any shade, you will need a panel at least 10 times as large. If you don't buy the highest quality selected cells, but go for most common panels, they have only 114 W/m2 nominal rated power, the required area will be 50% more.
If you would want to supply that 480 W round the clock, you will need batteries, account for the loss of power while charging (10% - 50% depending on what batteries and charge controllers you use), and take into account that sun only shines a small part of the day.
I read that from a panel in horizontal position, in London, UK, in December, you get on average 0.434 Wh of energy per each nominal W of panel power per day. In July the value is about 10 times as much Wh per day.
That means to provide 480 W continuously even in December, you would need a panel of nominal power 26544 W, which would be at least 154 m2 in size. And that's not taking into account loss on charging the battery.