Find the Wavy Words!

05AB1E, 11 9 bytes (Thanks to Adnan)

Dg2›iÇü‹Ù

Try it online!

Wavy Cases:

0 - Decreasing Wavy

1 - Increasing Wavy

Not Wavy Cases:

[0,1] - Not wavy, initially decreasing, but then has an increase/equality that broke the pattern.

[1,0] - Not wavy, initially increasing, but then has a decrease/equality that broke the pattern

Input String - Not possible to be wavy in the first place due to length.

Explanation:

Dg2›iÇü‹Ù   # Full program

D           # Push 2 copies of input.
 g2›i       # If length is greater than 2. 
     Ç      # Push ASCII values for all characters in the string.
      ü     # Push pairwise array.
       ‹    # Vectorize 1 if negative difference, 0 if positive difference.
        Ù   # Uniquify using right most unique values first.
            # Else just print the string back, letting them know it's not valid input.

Jelly, 10 bytes

OIṠḟ0µL’aQ

TryItOnline! or run all test cases

Returns:
[1] for wavy increasing
[-1] for wavy decreasing
something else otherwise ([], [0], [-1,1], or [1,-1])

(Declared as unnecessary: For a single value for each OIṠḟ0µL’aQS (11 bytes) will return 1, -1, and 0 respectively.)

How?

OIṠḟ0µL’aQ - Main link: s
O          - cast to ordinals
 I         - incremental differences
  Ṡ        - sign (-1 for decreasing, 0 for no change, 1 for increasing)
   ḟ0      - filter out zeros
     µ     - monadic chain separation
      L    - length
       ’   - decremented
        a  - and
         Q - unique items

Python 3, 77 75 bytes

lambda x:(len(set(x))>2)*(list(x)==sorted(x)or(list(x)==sorted(x)[::-1])*2)

Assumes all letters are of the same case.

Returns:

• 0 if not wavy
• 1 if forwards wavy
• 2 if backwards wavy

Removed unnecessary spaces thanks @ETHproductions