I am trying to understand the diode in parallel with Reset-Resistor. Is that because when the power is turned off 'VCC' becomes "GND" (0V), and the capacitor discharges through the diode bypassing the resistor? (Path of least resistance? Could I put an LED there to see this discharge?)
That's it. VCC becomes 0V, and without the diode the capacitor will try to discharge through the inverter's input protection (if it has any - 74HC does, I don't remember about 74LS).
An LED wouldn't do much good. If the inverter input has protection diodes, they'll keep the capacitor just above VCC as it heads down, and the LED won't light. If it doesn't, the LED will only discharge the capacitor to its forward voltage (a couple of volts or so), which is likely to be outside the safe range for the input.
I don't know if it's strictly necessary, but it doesn't hurt. It depends how tough the inverter's inputs are. I've always put the diode in, just to be safe.
It doesn't really matter which processor you use. If you come from an Intel world, the Z80 is likely to be a little easier on the software side. But anyone who survived the C64 vs Spectrum wars knows the 6502 is the best :-)
Quinn's work is great, and worth looking at whatever you use. She's much better at testing and step by step progress than I am. And it's all documented very well.
Edit: And it's fine to ignore the interrupt on the UART. You can start out with polling to see if it's ready for another character. Later on, if you decide you need it, you only need to add one wire. Probably. I say this without reading the datasheets, and never having used those particular chips.