Electronic – Why does voltage stay the same regardless of resistor connect to the source

Yes, the leakage current of D1 causes a voltage accross R1, which lowers the on-state voltage. Use a lower R1 or a better diode.

There are other Schottky diodes that have less leakage than 2 mA. Even this one probably has much lower leakage at 25°C. If you know it will always be operating at that temperature or less, then you can use the lower leakage figure. If it has to work at high temperatures, then select a better Schottky or perhaps resort to a ordinary silicon diode, like 1N4148 or equivalent. That's a nice and fast diode, but will have higher forward drop so therefore clip the undershoots at a lower level. Everything is a tradeoff.

From your description, I believe this is what you want to say (and be disciplined enough to tell the details properly).

This is on basic diode circuit analysis. For an ideal diode in this configuration, the load resistor has a zero voltage across it when the AC input is in the positive half cycle. Why is this the case? Your load is in parallel with a short.

What happens? You can look at it in two ways. First, electrical current takes the path of least resistance. It will only run through the short. No current means no voltage for an Ohmic resistor (remember Ohm's Law). And second, your load resistor is in a parallel configuration with a short. A voltage across a short is always zero volts. And in a parallel configuration, the voltage must be the same for each branch.

What is the consequence then for the negative half cycle?

Your diode is in a reverse bias configuration. It means your diode behaves like an open. There is a voltage drop across the load resistor because the current passes through the load resistor and not through the diode.

^{simulate this circuit – Schematic created using CircuitLab}