functions as applicative functors (Haskell / LYAH)

First of all, remember how fmap is defined for applicatives:

fmap f x = pure f <*> x

This means that your example is the same as (fmap (+ 5) (* 3)) 4. The fmap function for functions is just composition, so your exact expression is the same as ((+ 5) . (* 3)) 4.

Now, let's think about why the instance is written the way it is. What <*> does is essentially apply a function in the functor to a value in the functor. Specializing to (->) r, this means it applies a function returned by a function from r to a value returned by a function from r. A function that returns a function is just a function of two arguments. So the real question is this: how would you apply a function of two arguments (r and a, returning b) to a value a returned by a function from r?

The first thing to note is that you have to return a value of type (->) r which means the result also has to be a function from r. For reference, here is the <*> function:

f <*> g = \x -> f x (g x)

Since we want to return a function taking a value of type r, x :: r. The function we return has to have a type r -> b. How can we get a value of type b? Well, we have a function f :: r -> a -> b. Since r is going to be the argument of the result function, we get that for free. So now we have a function from a -> b. So, as long as we have some value of type a, we can get a value of type b. But how do we get a value of type a? Well, we have another function g :: r -> a. So we can take our value of type r (the parameter x) and use it to get a value of type a.

So the final idea is simple: we use the parameter to first get a value of type a by plugging it into g. The parameter has type r, g has type r -> a, so we have an a. Then, we plug both the parameter and the new value into f. We need both because f has a type r -> a -> b. Once we plug both an r and an a in, we have a b1. Since the parameter is in a lambda, the result has a type r -> b, which is what we want.

Going through your original question, I think there's one subtle but very key point that you might have missed. Using the original example from LYAH:

(+) <$> (+3) <*> (*100) $ 5

This is the same as:

pure (+) <*> (+3) <*> (*100) $ 5

The key here is the pure before (+), which has the effect of boxing (+) as an Applicative. If you look at how pure is defined, you can see that to unbox it, you need to provide an additional argument, which can be anything. Applying <*> to (+) <$> (+3), we get

\x -> (pure (+)) x ((+3) x)

Notice in (pure (+)) x, we are applying x to pure to unbox (+). So we now have

\x -> (+) ((+3) x)

Adding (*100) to get (+) <$> (+3) <*> (*100) and apply <*> again, we get

\y -> (\x -> (+) ((+3) x)) y ((*100) y) {Since f <*> g = f x (g x)}

5  -> (\x -> (+) ((+3) x)) 5 ((*100) 5)

(\x -> (+) ((+3) x)) 5 (500)

5 -> (+) ((+3) 5) (500)

(+) 8 500

508

So in conclusion, the x after f is NOT the first argument to our binary operator, it is used to UNBOX the operator inside pure.