The short answer to this is that, Python is a pass-by-object-reference language, not pass-by-reference as implied in the question. It means that:

1. result and result_tail are two variables that happen to point at the same value
2. Mutation / Changing of the underlying value (result_tail.next = ListNode(1)) will affect the value shown by result
3. However, assigning / pointing the variable result_tail to another value will NOT affect the value of result
4. result_tail = result_tail.next is assigning the next node of the node that is currently assigned by the variable

The following is an visualization of the values that are assigned to the variables (r = result, rt = result_tail):

result = ListNode(0)
#r
#0 -> None

result_tail = result
#r
#0 -> None
#rt

result_tail.next = ListNode(1)
#r
#0 -> 1 -> None
#rt

result_tail = result_tail.next
#r
#0 -> 1 -> None
#     rt

result_tail.next = ListNode(2)
#r
#0 -> 1 -> 2 -> None
#     rt

result_tail = result_tail.next
#r
#0 -> 1 -> 2 -> None
#          rt

References for additional reading:

• An article explaining the Python pass-by-object reference style in detail https://robertheaton.com/2014/02/09/pythons-pass-by-object-reference-as-explained-by-philip-k-dick/
• An answer explaining Python's pass-by-object reference style https://stackoverflow.com/a/33066581/12295149
• Question asking on Python's object reference style Understanding Python's call-by-object style of passing function arguments

To get the output you have listed at the end of your question, you would need to create a linked list. For instance, if you define the following class:

class ListNode:
    def __init__(self, val, nxt=None):
        self.val = val
        self.next = nxt

    def __repr__(self):
        return f"ListNode{{val: {self.val}, next: {self.next}}}"

And if you then define list1 as follows:

list1 = ListNode(1, ListNode(2, ListNode(4, None)))

Then the "commands" will give the output that you listed.

List to Linked List

If you want to create the above linked list from the list [1,2,4], then use this function:

def createLinkedList(values):
    head = None
    for val in reversed(values):
        head = ListNode(val, head)
    return head

Now you can convert a plain list to a linked list as follows:

list1 = createLinkedList([1,2,4])

Linked List to list

If you want to do the opposite, and create a standard list from a linked list, then define this function:

def linkedListIterator(head):
    while head:
        yield head.val
        head = head.next

Now, if you have a linked list, you can pass it to the above function. For instance:

list1 = createLinkedList([1,2,4])
lst = list(linkedListIterator(list1))

lst will be [1,2,4]