[yorwba]

  Prelude> y f = f (y f)
  Prelude> :type y
  y :: (t -> t) -> t

Because f is applied to an argument, it must be a function of generic type t -> u. Because y is applied to f, it must be a function of type (t -> u) -> v. Then the type of y f is v. Because f is applied to y f, t = v. Then the type of f (y f) is u. Because y f is defined as f (y f), the two expressions need to have the same type, t = u. Therefore f is of type t -> t and y is of type (t -> t) -> t.

[user2994cb]

It's true that the original non-recursive definition of Y doesn't typecheck in Haskell:

  Prelude> \f->(\x->f(x x))(\x->f(x x))

  <interactive>:5:14:
    Occurs check: cannot construct the infinite type: r0 ~ r0 -> t
    ...