You don't have to syntactically consume all of arguments of a function to write its body.

    f :: a -> Int -> Int -> Int
    f x = (+)

There really isn't anything meaningful to do with a function before it has consumed all it's arguments other than returning another function.

> However, this "f" is quite different from a function that has the signature "x->(y->z)"

Yes, we have applied the argument of type 'x'. f 0 is no longer of type "x -> y -> z", or "x -> (y -> z)" (because of -> precedence, the two are equivalent), it is "y -> z".

Currying is explicit.

    f x y = foobar

is just short hand for

    f = \x -> \y -> foobar