Electrical – power 12v fan using laptop charger while running laptop

1 - Look, the first thing you need to do is get a cheap DMM. As it stands you have no idea what is actually going on in your batteries, and frankly I'm surprised at how durable lead-acids can be when subjected to the abuse that you are apparently putting them through. Once you have a meter, you can look up "lead-acid charging algorithm" online and learn how to do this right.

2 - Your 48-volt charger was an abortion. Never, ever, use a charger that boils the electrolyte. I suspect that the only reason the PSUs don't cause boiling is that they simply can't produce enough power, while the monster can. And does. Get rid of it, or learn enough electronics to figure out the charging circuit and modify it. You are causing long-term damage to the batteries if you use it.

3 - There is no "intelligent coordination" going on. All of your PSUs are in current limit, and the smaller ones simply don't handle it as well as the larger. The exact hows and whys depend on the circuit design and components/assembly techniques, but it's not a matter of coordination, intelligent or otherwise.

^{simulate this circuit – Schematic created using CircuitLab}

You can power the MT3608 from a large capacitor on its input.

You will need a diode between the 5V supply and the capacitor to prevent the capacitor from back-powering the 5V supply once the 5V is off, . This could be either a schottky diode, or an ideal diode controller such as those made by Linear Tech.

Assuming you really get the 93% efficiency stated in the MT3608 datasheet, the input power will be 0.84W / 0.93 = 0.903W.

You say you want to power the fan for 5s, so the total energy is 5s * 0.903W = 4.515J.

The MT3608 has a 2V to 24V input range.

If the capacitor is to supply power to the fan, the energy lost by the capacitor will be equal to the energy input to the converter to power the fan, which is 4.515J.

The initial voltage on the capacitor is 5V minus a diode drop. Lets assume the initial voltage is 4.5V. The final voltage is 2V, which is the lower limit of the input range for the MT3608. So...

0.5 * C * ((4.5V)^2 - (2V)^2) = 4.515J

So C = 4.515J * 2 / ((4.5V)^2 - (2V)^2) = 555mF.

So you need a capacitor rated to at least 5V and 555mF.

You will need to size the capacitor larger than 555mF to account for any voltage drop on the output due to the internal ESR of the capacitor, as well as decreasing efficiency on the converter as the input voltage gets lower.