In lectures we said that if $X$ is independently uniformly distributed, the CLI statest that:
$$\lim_{n\rightarrow \infty} (\frac{\sum_{n=1}^{i}(x_i - \mu)}{\sqrt{n}\sigma}) =^{d} N(0,1)$$
However, I don't understand, what is $\sqrt{n}$ doing there, because we also did an example of binomial distribution (let's call it $S_n$), and we said: $$\lim_{n\rightarrow \infty} (\frac{S_n-np}{\sqrt{np(1-p)}}) =^{d} N(0,1)$$
From which I would conclude that the correct formula would be:
$$\lim_{n\rightarrow \infty} (\frac{Distr - expectation }{\sqrt{variance}}) =^{d} N(0,1)$$
What is correct and why?
The mean of $\sum_{i=1}^n x_i$ is $n \mu$ and its variance is $n\sigma^2$, so the central limit theorem reproduces your expectations ($\sqrt{\text{variance}}=\sqrt n \sigma$). All three equations you wrote down are correct.