Need to speed up my code.

[perfringo]

So the oneliner using collections.Counter seems to be fastest, but I cant understand why and how.
How to do this better with using just simple and readable basic python?

I believe that collections.Counter (and other stuff from built-in collections module) is worth of learning and should be considered as 'simple and readable basic Python'. This is special data structure for counting and it would be non-productive to invent a wheel. Just type help(collections.Counter) into interactive interpreter and you get description, examples etc.

If no Counter allowed or understood then it of course can be done 'manually' (and with O(n) - iterating only once over list):

>>> lst = [3,5,2,1,2,3,2,2]             # list to be iterated
>>> counts = dict()                     # dictionary for keeping counts of occurances
>>> for item in lst:                    # iterate all list items
...     try:                            # try +1 to value of key
...         counts[item] += 1
...     except KeyError:                # if key is not existing (raises KeyError)
...         counts[item] = 1            # add key and set value to 1 (first count)
... 
>>> counts[lst[-1]]                     # retrieve count of last item in lst from count dict
4

If try...except is not qualifying as 'simple readable and basic Python' then we can make it even simpler by not using 'advanced Python':

>>> counts = dict()
>>> for item in lst:
...     if item in counts:                  # if item already counted add one
...         counts[item] += 1
...     else:                               # else means that first time encountered
...         counts[item] = 1                # add key and set value to one
... 
>>> counts[lst[-1]]                         # get count of last item in lst
4
>>> counts                                  # whole counts dictionary
{3: 2, 5: 1, 2: 4, 1: 1}

There is also collections.defaultdict but in sense of 'basic Python' it's not any better than collections.Counter.

EDIT:

And of course there is always:

>>> lst = [3,5,2,1,2,3,2,2]                                                    
>>> lst.count(lst[-1])                                                            
4