Define value in __init__(). There is no other way to define instance attributes.

Attributes bound outside a instance method are class attributes and shared by all instances of that class. Hence modifications of the objects bound to class attributes affect all instances of the class, as you've noticed in your example.

A list comprehension can be used to fill a list with separate instances of a class, like so:

instancelist = [MyClass() for i in range(29)]

This avoids the problem with multiplying a list of one element with *, which re-uses the same object.

Python is a dynamic language so there is no concept of array of type.
You create an empty generic list with:

self.arrayOfGhosts = []

You don't care about the capacity of the list as it's dynamically allocated as well.
It's up to you to fill it with as many Ghost instances as you wish with:

self.arrayOfGhosts.append(Ghost())

The above is really enough, however:
If you really want to enforce this list to accept only Ghost and inheriting classes instances, you can create a custom list type like this:

class GhostList(list):

    def __init__(self, iterable=None):
        """Override initializer which can accept iterable"""
        super(GhostList, self).__init__()
        if iterable:
            for item in iterable:
                self.append(item)

    def append(self, item):
        if isinstance(item, Ghost):
            super(GhostList, self).append(item)
        else:
            raise ValueError('Ghosts allowed only')

    def insert(self, index, item):
        if isinstance(item, Ghost):
            super(GhostList, self).insert(index, item)
        else:
            raise ValueError('Ghosts allowed only')

    def __add__(self, item):
        if isinstance(item, Ghost):
            super(GhostList, self).__add__(item)
        else:
            raise ValueError('Ghosts allowed only')

    def __iadd__(self, item):
        if isinstance(item, Ghost):
            super(GhostList, self).__iadd__(item)
        else:
            raise ValueError('Ghosts allowed only')

Then for two-dimensional list you use this class like:

self.arrayOfGhosts = []
self.arrayOfGhosts.append(GhostList())
self.arrayOfGhosts[0].append(Ghost())

In Python, there is difference between __init__ and __new__. That is, by the time your code hits __init__ it is already "constructed". Only then you can add new "fields" to an instance, like for example self.myList = [] in __init__.

What def addSomething(cls, x) goes, you may named it cls but it is the instance in fact. So you should write self.myList.append() instead.

The problem lies in this line:

def __init__(self, lst=[], intg=0):

You shouldn't use a list as a default argument. The first time __init__ is called without lst specified the Python interpreter will define an empty list []. Subsequent calls to the function will operate on the same list if lst is not specified, without declaring a new list. This causes weird problems.

You should instead use a default value of None and add a check at the beginning of the function:

def __init__(self, lst=None, intg=0):
    if lst is None:
        lst = []

See this post for further details. Quoting the post:

Default arguments are evaluated at function definition time, so they're persistent across calls. This has some interesting (and confusing) side effects. An example:

>>> def foo(d=[]):
...     d.append('a')
...     return d

If you've not tried this before, you probably expect foo to always return ['a']: it should start with an empty list, append 'a' to it, and return. Here's what it actually does:

>>> foo() ['a']
>>> foo() ['a', 'a']
>>> foo() ['a', 'a', 'a']

This is because the default value for d is allocated when the function is created, not when it's called. Each time the function is called, the value is still hanging around from the last call. This gets even weirder if you throw threads into the mix. If two different threads are executing the function at the same time, and one of them changes a default argument, they both will see the change.

Of course, all of this is only true if the default argument's value is a mutable type. If we change foo to be defined as

>>> def foo2(d=0):
...     d += 1
...     return d

then it will always return 1. (The difference here is that in foo2, the variable d is being reassigned, while in foo its value was being changed.)