How can I count occurrences of sublists in a list and display them as dict

You could use a defaultdict of Counter to build the dictionary:

from collections import defaultdict, Counter

lst = [["Orange", "Carrot"], ["Green", "Apple"], ["Yellow", "Banana"], ["Orange", "Pumpkin"], ["Green", "Apple"]]

d = defaultdict(Counter)

for key, v in lst:
    d[key][v] += 1

res = {k: dict(v) for k, v in d.items()}
print(res)

Output

{'Orange': {'Carrot': 1, 'Pumpkin': 1}, 'Green': {'Apple': 2}, 'Yellow': {'Banana': 1}}

lst = [["Orange", "Carrot"], ["Green", "Apple"], ["Yellow", "Banana"], ["Orange", "Pumpkin"], ["Green", "Apple"]]
dict_sum = dict()

for item in lst:
    color = item[0]
    vegetable = item[1]
    
    # search in dict_sum
    if color not in dict_sum:
        dict_sum[color] = dict()
    
    # search vegetable in color
    if vegetable not in dict_sum[color]:
        dict_sum[color][vegetable] = 0

    # increase count
    dict_sum[color][vegetable] += 1
print(dict_sum)

Not an elegant solution but you can iterate over the list and create a dictionary:

In [14]: lst = [["Orange", "Carrot"], ["Green", "Apple"], ["Yellow", "Banana"], ["Orange", "Pumpkin"], ["Green", "Apple"]]                                                        

In [15]: k = {}                                                                                                                                                                   

In [16]: for elem in lst: 
    ...:     if k.get(elem[0],None): 
    ...:         nested = k.get(elem[0]) 
    ...:         if nested.get(elem[1],None): 
    ...:             nested[elem[1]] = nested.get(elem[1])+1 
    ...:         else: 
    ...:             nested[elem[1]]=1 
    ...:     else: 
    ...:         k[elem[0]] = {elem[1]:1} 


{'Orange': {'Carrot': 1, 'Pumpkin': 1},
 'Green': {'Apple': 2},
 'Yellow': {'Banana': 1}}