I had the same issue and had to do a bit more reading/thinking on it. Hopefully this will help!
Let’s simplify this down to the number of ways to get to 5 spaces. There are eight.
a)2+2+1 b)2+1+1+1 c)1+2+1+1 d)1+1+2+1 e)1+1+1+1+1 f)2+1+2 g)1+2+2 h)2+1+2
Now let’s look at the way to get to 4 spaces:
i)2+2 j)2+1+1 k)1+2+1 l)1+1+2 m)1+1+1+1
Finally, let’s look at all the ways to get 3 spaces: n)2+1 o)1+2 p)1+1+1
If you notice, all of the examples for getting 5 are actually ALL of the ways to get 3 and all of the ways to get 4 with just one additional step.
- a = i +1
- b = j + 1
- c = k + 1
- d = l + 1
- e = m +1
- f = n + 2
- g = o +2
- h = p +2
If you’ll notice, 1)-5) are all of the ways you can go to pad 4 with an additional + 1 and 6)-8) are all the ways you can get to pad 3 with an additional + 2. This is what they mean when they say that the only way you can get to the 5th lilypad is to either be at the 3rd lilypad and make a jump of 2 or to be at the 4th lilypad and make a jump of 1. Any way that you could go from pad 3 to pad 5 by first resting at 4 is already encapsulated by all the ways you can get to pad 5 via pad 4—because all of the ways to 4 necessarily include any permutation where the penultimate move is at 3. Trying to include again would be tantamount to double counting.
More concisely: The number of routes to X_n is the same as the number of routes to X_{n-2} plus the number of routes to X_{-1}—but the individual unique sequences are necessarily elongated by the addition of a +2 and +1, respectively.