Bob, thank you for your answer. This is starting to make sense. But do you mind elaborating a teeny bit more on why the total charge for the series case is Q not 2Q?

Because it seems to me that the overall effect of having two plates in series each with charge Q would give 2Q

@user10796158 Ok but I need some time to respond because I need to tend to something else at this time. Please stand by.

@user10796158 Look at the four plates in bottom left of the picture. You might be thinking that they get charges of Q, -Q, Q, -Q; this is not what happens, because plates 2 and 3 are connected with a conductor and no battery, so their charges have to be equal.

Am I correct in thinking that (perfect) capacitors don't flow current? As in, it all builds up on one plate inside the cap but doesn't cross to the other? And the only way current can flow out of a cap is back out the way it came in? And if all of that is correct, surely adding capacitors in series does absolutely nothing past the first cap? Because current doesn't flow any further than the first cap in the series? So the total capacitance would be that of the first in the line? Or do I completely misunderstand?

@Clonkex Yes, ideal capacitors don't allow current to pass through. However, what you'll still see is that just like a single capacitor, a series of capacitors has a positive charge on one end and negative charge on the other - the problem is that this is not true for the individual capacitors in the series; the "negative" plate of one capacitor is connected with a conductor to the "positive" plate of the succeeding capacitor, so their charges are the same. So the series is a capacitor, but not a "series of capacitors" - you're making one capacitor with a big gap, ergo less capacitance.

Note that (once we have the fundamental "why" out of the way), the formula for the capacitance of capacitors in series is exactly the same as the formula for the resistance of resistors in parallel.

@user10796158 Sorry for taking so long to get back (tied up with other things). I see others have addressed you edited post. Just in case you may still have doubts, I have added a somewhat different explanation to my answer. See the revision. Hope this helps clear things up for you.

@Clonkex, wrote "Because current doesn't flow any further than the first cap in the series?" - consider the following series connected components: ammeter - capacitor - ammeter - capacitor - ammeter. Next, connect this series combination to some external circuit, e.g., a voltage source in series with a resistor. It seems that you're claiming that only one of the ammeters will read a non-zero current but, in fact, all three ammeters will read the same non-zero current. Despite the fact that no charge flows from one plate to the other of a capacitor, there is nonetheless a current through.

@Luaan wrote the "negative" plate of one capacitor is connected with a conductor to the "positive" plate of the succeeding capacitor, so their charges are the same. - I don't believe that is correct. It's true that the two connected 'inner' plates have the same potential but that doesn't imply the plates have the same charge.

"When a second equal capacitor is connected in series with the first, the total battery voltage splits across each capacitor and is V/2 ... In other words, in contrast with putting the second capacitor in parallel with the first, the battery does not "charge" the second capacitor. The existing charge is simply redistributed." - this isn't correct is it? First, let me summarize what I think you're describing: (...)

(1) Initially there is just one capacitor with capacitance $C$ across the battery with voltage $V$ across and zero current through (DC steady state). This capacitor has charge $Q = CV$ on its 'top' plate and $-Q$ on its 'bottom' plate. (2) a second, identical (uncharged) capacitor is 'instantaneously' placed in series with the first. Now, it seems to me, there is no transient. The circuit is in a valid DC steady state solution with voltage $V$ across the first capacitor and zero voltage across the second capacitor and so no redistribution of charge takes place. (...)

I believe it's impossible for this circuit to 'end up' with voltage $V/2$ across each capacitor for this scenario. For there to be voltage $V/2$ across the second capacitor (in steady state), there must be a transient charging current 'down' through that capacitor. Since the capacitors are in series, they have identical current through. So, a current that charges the second capacitor to $V/2$ would also charge the first capacitor to voltage $3V/2$ which is inconsistent with KVL.

@AlfredCentauri Good points. For one thing, you can't change the voltage across a capacitor instantaneously. But do you agree if you connect two uncharged equal capacitors in series each will wind up with one half the charge that would one of the capacitors would have had if alone were placed across the same voltage source? For either case the total stored charge in the circuit will be the same.

@AlfredCentauri How is it possible for all three ammeters to read a current in your hypothetical situation? There are literally physical gaps in the circuit and the voltage is (presumably) not high enough to arc those gaps. How does current flow?

@Clonkex, the canonical explanation is to say that there is a displacement current (related to the changing electric field) between the plates of a capacitor that is charging / discharging.

@Conkex Re Alpha Centuri comment see Maxwells modification of Ampere’s law that adds displacement current

@svavil wrote ". You might be thinking that they get charges of Q, -Q, Q, -Q; this is not what happens" - I'm afraid that it is just plain wrong (despite 8 up-votes!). It's true that the inner plates have the same potential but it's not true that the have the same charge. From an electrostatics perspective, the charge distribution that minimizes the energy is not that the inner plates have the same charge but, rather, that the inner plates have the equal and opposite charge of their respective outer plates. There are plenty of references online that show this. (...)

@svavil, (...) Indeed, if it were the case that the inner plates have the same charge, the formula for the equivalent capacitance of series connected capacitors would be wrong since it is derived under the assumption that "they get charges of Q, -Q, Q, -Q", i.e., that the electric field due to the charge distribution is confined to the region between the plates of each individual capacitor.

@AlfredCentauri your reasoning looks correct to me, I think I'll retract my comment shortly.