Fill in same amount of characters where other column is NaN

One way is using str.repeat and fillna() not sure how efficient this is though:

df.Col2.fillna(pd.Series(['~']*len(df)).str.repeat(df.Col1.str.count(',')))

---

0       aa~bb~cc~dd
1               ~~~
2    ii~jj~kk~ll~mm
Name: Col2, dtype: object

pd.concat

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in df}, axis=1
).stack()

    Col1 Col2
0 0    a   aa
  1    b   bb
  2    c   cc
  3    d   dd
1 0    e  NaN
  1    f  NaN
  2    g  NaN
  3    h  NaN
2 0    i   ii
  1    j   jj
  2    k   kk
  3    l   ll
  4    m   mm

This loops on columns in df. It may be wiser to loop on keys in the delims dictionary.

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in delims}, axis=1
).stack()

---

Same thing, different look

delims = {'Col1': ',', 'Col2': '~'}
def f(c): return df[c].str.split(delims[c], expand=True)
pd.concat(map(f, delims), keys=delims, axis=1).stack()

zip_longest can be useful here, given you don't need the original Index. It will work regardless of which column has more splits:

from itertools import zip_longest, chain

df = pd.DataFrame({'Col1':['a,b,c,d', 'e,f,g,h', 'i,j,k,l,m', 'x,y'],
                   'Col2':['aa~bb~cc~dd', np.NaN, 'ii~jj~kk~ll~mm', 'xx~yy~zz']})
#        Col1            Col2
#0    a,b,c,d     aa~bb~cc~dd
#1    e,f,g,h             NaN
#2  i,j,k,l,m  ii~jj~kk~ll~mm
#3        x,y        xx~yy~zz

---

l = [zip_longest(*x, fillvalue='') 
     for x in zip(df.Col1.str.split(',').fillna(''), 
                  df.Col2.str.split('~').fillna(''))]

pd.DataFrame(chain.from_iterable(l))

    0   1
0   a  aa
1   b  bb
2   c  cc
3   d  dd
4   e    
5   f    
6   g    
7   h    
8   i  ii
9   j  jj
10  k  kk
11  l  ll
12  m  mm
13  x  xx
14  y  yy
15     zz