Showing that $S_n -\lfloor S_n \rfloor \sim U[0,1]$
This is one of those things that's annoying to compute directly, but becomes easy if you use modular arithmetic. In this case, you should work with real numbers modulo $1$. Then the claim follows directly from the fact that the uniform measure on $\Bbb R/\Bbb Z$ is invariant under convolution (i.e., independent sums).
Here is the argument drawn out in full detail, in case it is helpful.
The idea is to work on $\Bbb R/\Bbb Z$ instead of $\Bbb R$. Let $\pi: \Bbb R \to \Bbb R/\Bbb Z$ be the projection map $x \mapsto x \pmod 1$. Henceforth, whenever I refer to a "sum" it will be with respect to the additive group structure on $\Bbb R/\Bbb Z$.
Let $Y_i=\pi(X_i)$. Note that the $Y_i$ are uniformly distributed on $\Bbb R/\Bbb Z$ (i.e., they are distributed according to arclength measure if you view $\Bbb R/\Bbb Z$ as a circle, or Haar measure if you view it as a topological group).
Also note that any finite sum of independent uniformly distributed variables in $\Bbb R/\Bbb Z$ is still uniformly distributed on $\Bbb R/\Bbb Z$ (i.e., the arclength measure on the circle is invariant under convolution of measures).
Note also that $\pi$ is a group homomorphism, so we have that $\pi(S_n) = \sum_1^n Y_i$. We thus conclude from the previous paragraph that $\pi(S_n)$ is uniformly distributed on $\Bbb R/\Bbb Z$.
The final step is to note that $\pi$ is invariant under $f$, i.e., $\pi \circ f = \pi$. Thus $\pi(f(S_n))$ has a uniform distribution on $\Bbb R / \Bbb Z$. But $\pi$ is invertible if we restrict its domain to $[0,1)$. Moreover, $f(S_n)$ takes values in $[0,1)$.
The pushforward by $\pi^{-1}$ of the uniform measure on $\Bbb R/\Bbb Z$ is the uniform measure on $[0,1)$, so we conclude that $f(S_n)$ is uniformly distributed on $[0,1)$.
Here is a more elementary rendition of @shalop's answer. The point is that it all boils down to showing the following two claims:
Claim.
- If $U \sim \mathcal{U}[0,k]$ for some $k\in\mathbb{N}$, then $f(U) \sim \mathcal{U}[0,1]$.
- If $U \sim \mathcal{U}[0,1]$ and $a \in \mathbb{R}$, then $f(a+U) \sim \mathcal{U}[0,1]$.
Using this claim, we know that
$$f(X_n+a) = f(f(X_n) + a) \sim \mathcal{U}[0,1]$$
whenever $a \in \mathbb{R}$ and $X_n \sim \mathcal{U}[0,k]$ for some $k\in\mathbb{N}$. Then for any $r \in [0, 1)$, by the independence of $X_n$ and $S_{n-1}$,
\begin{align*} \mathbb{P}(Z_{n} \leq r) = \mathbb{E}[\mathbb{P}(f(X_{n}+S_{n-1}) \leq r \mid S_{n-1})] = \mathbb{E}[r] = r. \end{align*}
Therefore the desired conclusion follows.
Proof of Claim. In the first part, it is clear that $f(U)$ takes values only in $[0,1)$. Now for any $r \in [0,1]$, we have
$$ P(f(U) \leq r) = \sum_{q=0}^{k-1} P(q \leq U \leq q+r) = \sum_{q=0}^{k-1} \frac{r}{k} = r, $$
and therefore $f(U)$ has the desired distribution. In the second part, write $a = \lfloor a \rfloor + \langle a \rangle$, where $\langle a \rangle$ denotes the fractional part of $a$. Then for any $r \in [0,1)$,
\begin{align*} P(f(a+U) \leq r) &= P(\{ 0 \leq U < 1 - \langle a \rangle \} \cap \{ U+\langle a \rangle \leq r \}) \\ &\quad + P( \{ 1 - \langle a \rangle \leq U < 1 \} \cap \{ U+\langle a \rangle - 1 \leq r \}). \end{align*}
Considering the cases $r < \langle a \rangle$ and $r \geq \langle a \rangle$ separately, this can be easily computed as $r$, again proving that $f(a+U) \sim \mathcal{U}[0,1]$. $\square$