FYI, those heatsinks are going to get mighty hot in there...
Regarding the spark, it is unlikely to damage the supply, and it's simply the equalization of charge between the supply's own internal capacitors, and yours. It is, however, a transient to be respected: semiconductors do not take much current, or energy, before being destroyed. If there were a series diode or transistor between those capacitors, do not expect it to last long (unless of course it's designed to limit current, in which case the spark you get will be much milder).
We can calculate the peak current by assuming the capacitors have an equivalent circuit of an ideal capacitor in series with a resistor (ESR). Typical electrolytic capacitors of that size have on the order of 0.1 ohms. Thus at the instant we connect them together, we have 24V across one capacitor, 0V across the other, and the total 24V across both ESRs in series. This draws a current of I = V/R = 24V / 0.2Ω = 120A. This is a lot of current for the microscopic contact point, which instantly melts and vaporizes, producing a spark. The current decays according to an exponential RC curve, which for a total of some 1000s uF, has a time constant of 200s microseconds, delivering a total energy of about half a joule.
A joule isn't much energy, mechanically speaking (doing "12oz curls", i.e. lifting a beer can, is about 1.7J), but for a tiny semiconductor device it's a huge amount. Even large transistors are only rated for avalanche pulses on the order of 100s mJ (so, comparable to the energy stored in the capacitors). Small transistors and diodes may fail at only 10mJ, and most ICs less than that. (Typical electrostatic discharge (ESD) events are on the order of 10mJ.)
So in short, it's actually a justified concern, it's not much of a spark to us sweaty meat-sacks, but to a transistor it's best avoided, through proper wiring and design.
Tim