I thought of this as though the list A.b was marked static - then any modification of b through any instance of A modifies the shared static member
>>> class A:
... b = [1,]
...
>>> obj = A()
>>> obj.b.append(2)
>>> A.b
[1, 2]
>>> foo = A()
>>> foo.b
[1, 2]
>>>But what does that have to do with assignment?
>>> a = 3
>>> a += 4
>>> print(a)
7
Obviously, this is inspired by the C += operator, and behaves accordingly, when it's not being weird.
The difference is in how the Class attributes are affected i.e. the values of "A.a" and "A.b".
# obj.a += (2,) has no impact on A.a since tuples
# are immutable and copies are made.
# This is unexpected from a C/C++ intuition.
>>> print(A.a)
(1,)
# obj.b += [2,] has an indirect impact on A.b
# since lists are mutable and updates are made in place.
# This is unexpected for beginners in Python since it
# deviates from how int, str and tuples behave as "+="
# can be used for int, str, tuples and lists.
>>> print(A.b)
[1, 2]
No, they don't.
obj.b does not exist, and no assignment to it takes place, and there is no “indirect effect”.
A.b exists, and because class attributes can be accessed as if they were members on instances, and because A.b has an implementation of in-place addition, that implementation is called on A.b instead of an assignment to obj.b. That is the only thing that happens as a result of the statement that looks like it might be an assignment to obj.b. The effect on A.b is the only action, not an indirect effect of an assignment.
Side note: the object can be immutable, but then the (variable or attribute) left side should be replaced with a new immutable object.