>>

>>First (i += 2)  sets i = 3
>>Second (i += 2)  sets i = 5
>>
>>Then
>>j = 2 + 3 + 5
>>

>
>Yes, I figured that, but more logically it would be
>
>j = 2 + 5 + 5

No,

it adds 2, then uses that value

On a stack we would have

2
3    // 2 is added to i, value of i = 3 here
5   //  2 is added to i, value of i = 5 here

Then two additions which would leave 10 on the stack