Thanks to all for weighing in. A key point to my understanding has come about with learning that electrons in a closed circuit flow opposite the arrow symbol on the transistor.
So wiith NAND 2 NPN transistor circuit with LED and 2 normally open switches, cathode is to ground with anode connected to collector/voltage in (via resistor). So the circuit is already closed, and the LED is on. The open switches have no bearing, nor does 1 closed switch.
However, when both switches are closed, a flood of electrons goes to the LED anode, essentially creating 2 grounds on the LED and the light goes off.
My previous confusion about why the output LED was "above" or "below" the transistors had to do with not understanding the difference of connecting the output to the collector vs. the emitter: unlike the NAND circuit where voltage is already being supplied to the output/LED, when the output is connected to emitter, no voltage is flowing through the transistor when both bases are open.