From what I understand, the voltage doubles since the reflected and incident pulses add up. But then why don't all open terminations have double the source voltage since we can consider the voltage there to be a pulse of infinite duration?
Yes, the voltage doubles when the pulse hits the open end of the cable, but this change then propagates back toward the source end of the cable. What happens there depends on how well the impedance of the source is matched to the impedance of the cable.
If the source is matched to the cable, the initial pulse was only half of the nominal source voltage to begin with, because of the voltage divider action between the source impedance and the cable impedance. When this pulse gets to the open end of the cable, it doubles to match the orignal source voltage. Once this change propagates back to the source, then the entire cable is at the nominal source voltage, and no further reflections occur.
If the source has a very low impedance with respect to the cable, then the backward propagating pulse will be reflected there, too, but now it will be inverted, but slightly attenuated. It will propagate once again to the open end of the cable, where the whole thing repeats. You'll see a decaying oscillatory waveform on the cable until all of the reflections die out — which happens quite quickly on a short cable. Eventually, the cable settles down at the nominal source voltage.
That's because a "pulse of infinite duration" is not at all similar to "a pulse of finite duration".
For the reflection, and all what comes from that, to be significant you have to be working with a time base that is somewhat similar to the propagation delay of the cable. Since the latter is generally in the 10ns ballpark you can easily see why that won't work.
When you turn on a lab PSU you probably have a spike at its terminals but the cables inside are so short that it is barely noticeable.