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

The problem is that you are missing the reference to the head of your list, because you are overwriting it. Start with this instead:

list1 = [4,5,1,2,0,4]
head = ListNode(list1[0])
tail = head

Then tail will be a reference to the last element of the linked list. Now in your loop you do:

while e < len(list1):
      print(head)
      tail.next = ListNode(list1[e])
      tail = tail.next
      e+=1

So you add an element to the list as before, but now we are modifying the tail variable. At the end:

return head

You will now be returning the head node of the list.