It is too long for a comment , Thus I wrote an answer

I am the person who wrote the transform that you mentioned in your question. I did not see anywhere the term of endless transform, I just noticed if we apply the variable transform
$$y=\frac{f(x)}{\frac{f'(x)}{2f(x)}+y_1} $$ on Riccati Differential Equation $y'+y^{2}=f(x)$,

We get same type Riccati Differantial Equation after the transform.
$$y_1 '+y_1^{2}=f_1(x)=f(x)+(\frac{-f'(x)}{2f(x)})^{2}+(\frac{-f'(x)}{2f(x)})'$$

Then, if we apply the transform $$y_1=\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+y_2} $$ on $y_1 '+y_1^{2}=f_1(x)$ , we get the same type equation again $$y_2 '+y_2^{2}=f_2(x)=f_1(x)+(\frac{-f_1'(x)}{2f_1(x)})^{2}+(\frac{-f_1'(x)}{2f_1(x)})'$$
It never ends , we always get the same type equation ($y'+y^{2}=f(x)$) , Thus I wrote it as endless transform.

Finally, the particular solution can be written in infinite terms as

$$y_p(x)=\frac{f(x)}{\frac{f'(x)}{2f(x)}+\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+\frac{f_2(x)}{\frac{f_2'(x)}{2f_2(x)}+.....}} } $$

Where $$f_{n+1}(x)=f_n(x)+(\frac{-f_n'(x)}{2f_n(x)})^{2}+(\frac{-f_n'(x)}{2f_n(x)})'$$
and $$f_0(x)=f(x)$$

I do not know the name of this transform in literature, So I named it as endless transform. Maybe someone else can help to find what the name of the transform is in literature.

I hope it helps to enlighten you

unendlich' = infinite was translated as 'endless'.) Looking at what was written, and allowing for the fact this is not an area of research for me, it looks like the OP means to find a fixed point $f$ of a differential or integral operator by means of a convergent infinite sequence in a function space, or something like that. Googling for "endless" something or other is almost certainly hopeless. Why do you really need to know? $\endgroup$ – Todd Trimble♦ Jul 5 '16 at 14:10