Make a loop within for loop statement in Python

Although,I couldnt figure out the reason why the s in loop isnt iterating; i though noticed another problem in the code i.e. everytime you set value of s as s+(i+2) eg 0+(1+2), the other time its value is changes as 3+(i+2) in place of 0+(i+2)

Though i have an alternate solution for your question i.e.

vbn = [1,0,2,3,0,4,5,0]
s = 0
arr = []*10 #list to store no. of positions we need to add 0 
for i,j in enumerate(vbn):
  if j == 0:
    arr.append(i)
m=0 #variable to keep count of number of times the list is extended before the current one
for k in arr:
  k+=m #since, when we add zero to the list the current positions of elements changes
  vbn.insert(k+1,0)
  m+=1
print(vbn)

You can try to create a new list and insert the element (the non-zero number or two zeros) in vbn to the new list.

vbn = [1,0,2,3,0,4,5,0]
newlist = []
for i in vbn: 
    if i == 0: 
        newlist.insert(len(newlist),i)
        newlist.insert(len(newlist),0)
    else: 
        newlist.insert(len(newlist),i)   
print (newlist)

The displayed results is:

[1, 0, 0, 2, 3, 0, 0, 4, 5, 0, 0]

As an alternative method if you don't want to iterate multiple values is to create a new array in a loop:

vbn = [1,0,2,3,0,4,5,0]
new = []

for i in vbn:
    if i == 0:
        new += [i, 0]
    else:
        new += [i]
print(new)

Your problem is that the indexes at which you're inserting are no longer valid after the first insert, because you have increased the size of the list. If you change your code to insert x instead of 0, and print vbn in each iteration of the loop, you can see what is happening:

vbn = [1,0,2,3,0,4,5,0]
s = 0
for i,j in enumerate(vbn[s:]):
    if j == 0:
        vbn.insert(i+1,'x')
        s += i+2
    print(vbn)

Output:

[1, 0, 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 'x', 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 'x', 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 'x', 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 'x', 0, 4, 'x', 5, 0]

As you can see, even the second insertion occurs at the wrong place. You can work around this by adding an offset to the insertion point, and increasing that offset each time you make an insert:

vbn = [1,0,2,3,0,4,5,0]
s = 0
o = 1
for i,j in enumerate(vbn[s:]):
    if j == 0:
        vbn.insert(i+o,'x')
        s += i+2
        o += 1
    print(vbn)

Output (in this case you can see x being inserted in the places you expect):

[1, 0, 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 'x', 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 'x', 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 'x', 4, 5, 0]
[1, 0, 'x', 2, 3, 0, 'x', 4, 5, 0, 'x']

Just change the x back to a 0 to get your desired result. Note that I'm not sure what the s += i+2 code is for; I've left it in on the presumption you use it after the loop.