super() and order of running method in class inheritance

[Gribouillis]

how does Python recognize this argument for super() in that line?

It is a very good question. I don't know the complete answer however the documentation of super() gives this example (with this exact comment)

class C(B):
    def method(self, arg):
        super().method(arg)    # This does the same thing as:
                               # super(C, self).method(arg)

Furthermore, if you type help(super) in the Python console, it says

class super(object)
 |  super() -> same as super(__class__, <first argument>)
 |  super(type) -> unbound super object
 |  super(type, obj) -> bound super object; requires isinstance(obj, type)
 |  super(type, type2) -> bound super object; requires issubclass(type2, type)
...

this suggests that it uses a local hidden variable named __class__. This variable can indeed be accessed in a method

>>> class A:
...     def foo(self):
...         print(__class__)
... 
>>> A().foo()
<class '__main__.A'>
>>> 

Further investigation indicates that this variable is loaded from a place called 'fast locals storage'

>>> import dis
>>> dis.dis(A.foo)
  3           0 LOAD_GLOBAL              0 (print)
              2 LOAD_DEREF               0 (__class__)
              4 CALL_FUNCTION            1
              6 POP_TOP
              8 LOAD_CONST               0 (None)
             10 RETURN_VALUE
>>> 

LOAD_DEREF(i)
Loads the cell contained in slot i of the “fast locals” storage. Pushes a reference to the object the cell contains on the stack.