If-statements are for the weak :-)

This uses numpy, but it's the shortest answer presented thus far.

    In [1]: from itertools import chain, zip_longest
       ...: import numpy as np
       ...:
       ...: def f(l):
       ...:     a = np.array([-np.inf, *l])
       ...:     up = a[:-1] <= a[1:]
       ...:
       ...:     pos = (up[1:] != up[:-1]).nonzero()[0]
       ...:     pos = [0, *(1+pos), len(l)]
       ...:
       ...:     up_runs = (l[x:y] for x,y in zip(pos[0::2], pos[1::2]))
       ...:     down_runs = (l[y-1:x-1:-1] for x,y in zip(pos[1::2], pos[2::2]))
       ...:
       ...:     pairs = zip_longest(up_runs, down_runs)
       ...:     return [*filter(None, chain(*pairs))]
       ...:
       ...: f([1, 2, 3, 2, 1, 4, 5, 6, 7])
    Out[1]: [[1, 2, 3], [1, 2], [4, 5, 6, 7]]