Because "append" existed long before "pop" was thought of. Python 0.9.1 supported list.append in early 1991. By comparison, here's part of a discussion on comp.lang.python about adding pop in 1997. Guido wrote:

To implement a stack, one would need to add a list.pop() primitive (and no, I'm not against this particular one on the basis of any principle). list.push() could be added for symmetry with list.pop() but I'm not a big fan of multiple names for the same operation -- sooner or later you're going to read code that uses the other one, so you need to learn both, which is more cognitive load.

You can also see he discusses the idea of if push/pop/put/pull should be at element [0] or after element [-1] where he posts a reference to Icon's list:

I stil think that all this is best left out of the list object implementation -- if you need a stack, or a queue, with particular semantics, write a little class that uses a lists

In other words, for stacks implemented directly as Python lists, which already supports fast append(), and del list[-1], it makes sense that list.pop() work by default on the last element. Even if other languages do it differently.

Implicit here is that most people need to append to a list, but many fewer have occasion to treat lists as stacks, which is why list.append came in so much earlier.

Your code works fine with a correct implementation of Stack and the expected sequence of characters is printed.

I think that your problem is only that the results are printed in modify_stack() and not being returned as the function return value. This is suggested by the way that function is supposed to be called - print(modify_stack('EAS*Y*QUE***ST***IO*N***')). It is clear that a return value from modify_stack() is expected.

Simply change your function to accumulate a list of the produced letter sequence, and then return that sequence as a string... like this:

class Stack(list):
    push = list.append

def modify_stack(symbolString):
    stack=Stack()
    result = []
    for character in symbolString:
        if character != '*':
            stack.push(character)
        else:
            result.append(stack.pop())
    return ''.join(result)

Now each popped character is added to the end of the result list whenever a * is seen. Finally the characters in result are converted to a string using join() and returned from the function

>>> print(modify_stack('EAS*Y*QUE***ST***IO*N***'))
SYEUQTSAONIE