python a,b = b,a implementation? How is it different from C++ swap function?

To understand this you need to go inside the implementation using dis.

First lets consider a simple swap function:

from dis import dis

def swap(i, j):
    i, j = j, i

dis(swap)

Output Byte Code:

4             0 LOAD_FAST                1 (j)
              2 LOAD_FAST                0 (i)
              4 ROT_TWO
              6 STORE_FAST               0 (i)
              8 STORE_FAST               1 (j)
             10 LOAD_CONST               0 (None)
             12 RETURN_VALUE

You can see that there is ROT_TWO which means that

Swaps the two top-most stack items.

This ROT_TWO is mainly responsible for the swapping.

Now coming to your question:

Lets take the example which is working:

from dis import dis

def swap():
    a = [2, 1]
    a[0], a[1] = a[1], a[0]

dis(swap)

Output Byte Code :

  4           0 LOAD_CONST               1 (2)
              2 LOAD_CONST               2 (1)
              4 BUILD_LIST               2
              6 STORE_FAST               0 (a)

  5           8 LOAD_FAST                0 (a)
             10 LOAD_CONST               2 (1)
             12 BINARY_SUBSCR
             14 LOAD_FAST                0 (a)
             16 LOAD_CONST               3 (0)
             18 BINARY_SUBSCR
             20 ROT_TWO
             22 LOAD_FAST                0 (a)
             24 LOAD_CONST               3 (0)
             26 STORE_SUBSCR
             28 LOAD_FAST                0 (a)
             30 LOAD_CONST               2 (1)
             32 STORE_SUBSCR
             34 LOAD_CONST               0 (None)
             36 RETURN_VALUE

Output byte code is similar to what we have when it is a simple swap function.

But when the code is changed:

from dis import dis

def swap():
    a = [1, 0]
    a[0], a[a[0]] = a[a[0]], a[0]
dis(swap)

swap()

Output is

  4           0 LOAD_CONST               1 (1)
              2 LOAD_CONST               2 (0)
              4 BUILD_LIST               2
              6 STORE_FAST               0 (a)

  5           8 LOAD_FAST                0 (a)
             10 LOAD_FAST                0 (a)
             12 LOAD_CONST               2 (0)
             14 BINARY_SUBSCR
             16 BINARY_SUBSCR
             18 LOAD_FAST                0 (a)
             20 LOAD_CONST               2 (0)
             22 BINARY_SUBSCR
             24 ROT_TWO
             26 LOAD_FAST                0 (a)
             28 LOAD_CONST               2 (0)
             30 STORE_SUBSCR
             32 LOAD_FAST                0 (a)
             34 LOAD_FAST                0 (a)
             36 LOAD_CONST               2 (0)
             38 BINARY_SUBSCR
             40 STORE_SUBSCR
             42 LOAD_CONST               0 (None)
             44 RETURN_VALUE

You can see the output byte code that top two items are the same. Hence it doesn't swap

This kind of the behaviour is indeed related to the way Python evaluates the expression of the type

a,b=b,a

In fact, what Python does is first it "prepares" the values of the right side by creating a tuple (b,a). Then this tuple is unpacked and assigned to the variables in the reverse order.

It is important to note that although Python uses references to objects the objects the variable names refer to may change if they refer to values of immutable type. It is not so with mutable types (illustrated by example in Python FAQ).

To break down the example with mutable types (lists) that you used:

a = [2,1,0]    
a[0], a[a[0]] = a[a[0]], a[0]

1. a[a[0]] takes the value from the a[0] element (equal to 2) of the list a (value 0).
2. a[0] is 2 hence the tuple created is (0,2)
3. Tuple (0,2) is unpacked and 0 replaces 2 in the list (0th element).
4. Now, a[a[0]] can be read as: take 0th element of list a (which is currently 0) and then replace the value in the list at that place with 2 from tuple unpacking (now 0 is replaced by 2 - which make the operation look like it does nothing to the list).

As suggested in the answer from von Oak changing the order helps because the step from the point 4. above does not replace the value again.

I suggest you refer to passing by assignment answer to understand functions and parameter passing.