as I understand it, mostly from the docs at the online "try it out" page:
from the warning given, if you pop the stack and the stack is empty, it will instead eval input. From the first answer you linked there I can see that this is cached and only takes input the first time it happens.
Then, something like this happens
h → push(pop() + 1)
Z → push(0)
R → a = pop(); b = pop(); push(a); push(b)
V → (according to the docs) count = pop(); while (count--) push(eval(pop()))
(or according to how it behaved in the interpreter earlier) count = pop(); item = pop(); while (count--) push(item);