The characteristic function is an expectation:
$$
\varphi_S(t) = \mathbb{E}\left(\exp(i S t)\right) = \mathbb{E}\left(\exp\left(i \left(U_1 + U_2 + \cdots + U_n \right) t\right)\right)
$$
Now, if $U_i$ is independent, the expectation factors into product of expectations, because :
$$
\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot \mathrm{e}^{i t U_2} \cdots \mathrm{e}^{i t U_n}\right) = \mathbb{E}\left(\mathrm{e}^{i t U_1} \right)\cdot \mathbb{E}\left(\mathrm{e}^{i t U_2} \right) \cdots \mathbb{E}\left(\mathrm{e}^{i t U_n}\right)
$$
Since $U_i$ have identical distributions, these expectations are the same, thus:
$$
\varphi_S(t) = \left(\mathbb{E}\left(\mathrm{e}^{i t U} \right)\right)^n = \left( \frac{\sin(t)}{t}\right)^n
$$
According to the inversion formula, there is one-to-one correspondence between cumulative distribution function, and the characteristic function. Since $\varphi_S(t)$ is a characteristic function for all $n \in \mathbb{N}$ by construction, the inversion formula applies for all natural $n$.
Moverover, if $\varphi_S(t)$ is integrable, then $F_S(x)$ is absolutely continuous, i.e. the notion of the probability density function is well-defined. Notice that $\varphi_S(t)$ is absolutely integrable for $n \geqslant 2$, by virtue of $|\varphi_S(t)| \leqslant \min\left(1, t^{-n}\right)$, and hence integrable. However, $\varphi_S(t)$ is also conditionally integrable for $n=1$ (the integral is known as Dirichlet integral).
Applying the inversion formula to obtain an explicit expression for the probability density is somewhat involved. This earlier post of mine contains the derivation.
To conclude, let me remark, that the sum of iid uniform continuous random variable follows what is known as the Irwin-Hall distribution.
