Is this lambda abstraction created as a generator of a recursive function?

In lambda calculus, a recursive function $f$ is obtained by

$$ f = Y g $$

where $Y$ is the Y combinator and $g$ is the generator of $f$ i.e. $f$ is a fixed point of $g$ i.e. $f == g f$.

In The Scheme Programming Language, I saw an example implementing a recursive function $f$ that sums the integers in a list:

(let ([sum (lambda (f ls)
                  (if (null? ls)
                     0
                     (+ (car ls) (f f (cdr ls)))))])
(sum sum '(1 2 3 4 5)))  => 15

What is the mathematical derivation that drives to create the lambda abstraction

lambda (f ls)
     (if (null? ls)
         0
        (+ (car ls) (f f (cdr ls))))

?
It looks like a generator of $f$, but not entirely.

Thanks.

If you look at the $Z$ combinator (because we're in an eager context), you have $$Z=\lambda g.(\lambda f.g(\lambda v.ffv))(\lambda f.g(\lambda v.ffv))$$

If you look at the code, we don't apply a fixed point combinator to sum, we simply apply sum to itself, so there's no reason for sum to be "generator".

You can also readily show that sum is (modulo currying) equal to $\lambda f.g(\lambda v.ffv)$ for some $g$, basically by replacing (f f ...) with (h ...) where h is the argument to $g$ in the definition for sum (minus the outermost lambda).

StackExchange Computer Science Q#113177, answer score: 4