Hint 1: You might want to study the Python code in the bisect module.

Hint 2: Slicing can be used for list insertion:

>>> s = ['a', 'b', 'd', 'e']
>>> s[2:2] = ['c']
>>> s
['a', 'b', 'c', 'd', 'e']

Being NumPy tagged and since input is mentioned as list/array, you can simply use builtin numpy.insert -

np.insert(original_list, insertion_indices, new_numbers)

To roll out the theory as a custom made one (mostly for performance), we could use mask, like so -

def insert_numbers(original_list,insertion_indices, new_numbers):
    # Length of output array                               
    n = len(original_list)+len(insertion_indices)

    # Setup mask array to selecrt between new and old numbers
    mask = np.ones(n,dtype=bool)
    mask[insertion_indices+np.arange(len(insertion_indices))] = 0

    # Setup output array for assigning values from old and new lists/arrays
    # by using mask and inverted mask version
    out = np.empty(n,dtype=int)
    out[mask] = original_list
    out[~mask] = new_numbers
    return out

For list output, append .tolist().

Sample run -

In [83]: original_list = [0, 1, 2, 3, 4, 5, 6, 7]
    ...: insertion_indices = [1, 4, 5]
    ...: new_numbers = [8, 9, 10]
    ...: 

In [85]: np.insert(original_list, insertion_indices, new_numbers)
Out[85]: array([ 0,  8,  1,  2,  3,  9,  4, 10,  5,  6,  7])

In [86]: np.insert(original_list, insertion_indices, new_numbers).tolist()
Out[86]: [0, 8, 1, 2, 3, 9, 4, 10, 5, 6, 7]

Runtime test on a 10000x scaled dataset -

In [184]: original_list = range(70000)
     ...: insertion_indices = np.sort(np.random.choice(len(original_list), 30000, replace=0)).tolist()
     ...: new_numbers = np.random.randint(0,10, len(insertion_indices)).tolist()
     ...: out1 = np.insert(original_list, insertion_indices, new_numbers)
     ...: out2 = insert_numbers(original_list, insertion_indices, new_numbers)
     ...: print np.allclose(out1, out2)
True

In [185]: %timeit np.insert(original_list, insertion_indices, new_numbers)
100 loops, best of 3: 5.37 ms per loop

In [186]: %timeit insert_numbers(original_list, insertion_indices, new_numbers)
100 loops, best of 3: 4.8 ms per loop

Let's test out with arrays as inputs -

In [190]: original_list = np.arange(70000)
     ...: insertion_indices = np.sort(np.random.choice(len(original_list), 30000, replace=0))
     ...: new_numbers = np.random.randint(0,10, len(insertion_indices))
     ...: out1 = np.insert(original_list, insertion_indices, new_numbers)
     ...: out2 = insert_numbers(original_list, insertion_indices, new_numbers)
     ...: print np.allclose(out1, out2)
True

In [191]: %timeit np.insert(original_list, insertion_indices, new_numbers)
1000 loops, best of 3: 1.48 ms per loop

In [192]: %timeit insert_numbers(original_list, insertion_indices, new_numbers)
1000 loops, best of 3: 1.07 ms per loop

The performance just shoots up, because there's no runtime overhead on conversion to list.