It's not clear to me why this specific task requires a loop:

array = [9, 8, 2, 3, 8, 3, 5]

def remove_2nd_occurance(array, value):

    ''' Raises ValueError if either of the two values aren't present '''

    array.pop(array.index(value, array.index(value) + 1))


remove_2nd_occurance(array, 8)

print(array)

Here's a way you can do this using itertools.count along with a generator:

from itertools import count

def get_nth_index(lst, item, n):
    c = count(1)
    return next((i for i, x in enumerate(lst) if x == item and next(c) == n), None)

a = [9,8,2,3,8,3,5]  
indx = get_nth_index(a, 8, 2)
if indx is not None:
    del a[indx]

print(a)
# [9, 8, 2, 3, 3, 5]

remove() removes the first item from the list which matches the specified value. To remove the second occurrence, you can use del instead of remove.The code should be simple to understand, I have used count to keep track of the number of occurrences of item and when count becomes 2, the element is deleted.

a = [9,8,2,3,8,3,5]
item  = 8
count = 0
for i in range(0,len(a)-1):
        if(item == a[i]):
               count =  count + 1
               if(count == 2):
                      del a[i]
                      break
print(a)

This one if you need to delete second and following occurrence of a target item:

# deleting second and following occurrence of target item
a = [9,8,2,3,8,3,5]
b = []
target = 8 # target item
for i in a:
    if i not in b or i != target:
        b.append(i)
a=b
print(a)
# [9, 8, 2, 3, 3, 5]

If you need to delete second and following occurrence for any item:

# deleting any second and following occurence of each item
a = [9,8,2,3,8,3,5]
b = []
for i in a:
    if i not in b:
        b.append(i)
a=b
print(a)
# [9, 8, 2, 3, 5]

Now, when you need to delete only second occurrence of target item:

# deleting specific occurence of target item only (use parameters below)
a = [9,8,2,3,8,3,5,8,8,8]
b = []

# set parameters
target = 8 # target item
occurence = 2 # occurence order number to delete

for i in a:
    if i == target and occurence-1 == 0:
        occurence = occurence-1
        continue
    elif i == target and occurence-1 != 0:
        occurence = occurence-1
        b.append(i)
    else:
        b.append(i)
a=b
print(a)
# [9, 8, 2, 8, 3, 5, 8, 8, 8]

python code to remove 2nd occurrence

ls = [1, 2, 3, 2]
index = ls.index(2, 0)
lss = ls[index + 1:]
lss.remove(2)
print(ls[:index + 1]+lss)

To remove 2nd Occurrence in list

list1=["L",1,"L",2,"L"]

print(list1)

print(list1.index("L",1))

list1.pop(2)

print(list1)