Building off of Alex's nice observation, I think I can prove:
\\( f(\frac{a}{b}) = \frac{b}{c} \Rightarrow f(f(\frac{a}{a + b})) = \frac{b}{b + c}\\).

First let's introduce an alternate definition of \\(f\\) that I think is a
little easier to work with. Recall that \\(f\\) is originally defined as:

$$f(x) = \frac{1}{2\lfloor x \rfloor - x + 1}.$$

The floor function in the denominator is kind of annoying, so let's
see if we can replace it with something that's easier (for me, at
least!) to reason about. The floor of x/y is just x/y minus its
fractional part, or formally:

$$\lfloor \frac{x}{y} \rfloor = \frac{x}{y} - \frac{x\mod y}{y}.$$

Plugging this in and rearranging, we get:
$$f(\frac{x}{y}) = \frac{y}{x + y - 2(x \mod y)}.$$

Now let's prove the claim. Suppose:
$$f(\frac{a}{b}) = \frac{b}{a + b - 2(a \mod b)} \equiv \frac{b}{c}.$$

Now $$\begin{align}f(\frac{a}{a + b}) &= \frac{a + b}{a + (a + b) - 2(a \mod (a + b))}\\ &= \frac{a + b}{b},\end{align},$$

and by applying \\(f\\) to both sides, we get:

$$f(f(\frac{a}{a + b}) = f(\frac{a + b}{b}) = \frac{b}{a + b + b - 2 ((a + b) \mod b)} = \frac{b}{a + b + b - 2 (a \mod b)}.$$

But by assumption, \\(c = a + b - 2(a \mod b)\\). Plugging this in, we obtain: \\(f(f(\frac{a}{a + b})) = \frac{b}{b + c}\\) as desired.