TL;DR

The below code is the fastest answer I could figure out from this question:

it = iter(df['a'])
df['b'] = [next(it).join(i) for i in df['b']]

The above code first does a generator of the a column, then you can use next for getting the next value every time, then in the list comprehension it joins the two strings.

Long answer:

Going to show my solutions:

Solution 1:

To use a list comprehension and a generator:

it = iter(df['a'])
df['b'] = [next(it).join(i) for i in df['b']]
print(df)

Solution 2:

Group by the index, and apply and str.join the two columns' value:

df['b'] = df.groupby(df.index).apply(lambda x: x['a'].item().join(x['b'].item()))
print(df)

Solution 3:

Use a list comprehension that iterates through both columns and str.joins:

df['b'] = [x.join(y) for x, y in df.values.tolist()]
print(df)

These codes all output:

   a          b
0  a  haealalao
1  b    gbobobd
2  c  gcrcecact
3  d    ndidcde

Timing:

Now it's time to move on to timing with the timeit module, here is the code we use to time:

from timeit import timeit
df = pd.DataFrame({'a': ['a', 'b', 'c', 'd'], 'b': ['hello', 'good', 'great', 'nice']})
def u11_1():
    it = iter(df['a'])
    df['b'] = [next(it).join(i) for i in df['b']]
    
def u11_2():
    df['b'] = df.groupby(df.index).apply(lambda x: x['a'].item().join(x['b'].item()))
    
def u11_3():
    df['b'] = [x.join(y) for x, y in df.values.tolist()]

print('Solution 1:', timeit(u11_1, number=5))
print('Solution 2:', timeit(u11_2, number=5))
print('Solution 3:', timeit(u11_3, number=5))

Output:

Solution 1: 0.007374127670871819
Solution 2: 0.05485127553865618
Solution 3: 0.05787154087587698

So the first solution is the quickest, using a generator.