Intuition behind $\phi(n) = \sum\limits_{d\mid n}\mu(d)\frac{n}{d}$

Yes, this is a correct and direct application of the inclusion-exclusion principle.

To be more explicit, without looking at the prime factorization :

$$\phi(n) = \sum_{\substack{m \le n\\ gcd(m,n)=1} }m^{0}=\sum_{m \le n} m^0 \sum_{d \ | \ gcd(m,n)} \mu(d) = \sum_d \mu(d) \sum_{\substack{m \le n\\ d \ |\ gcd(m,n)}}m^{0} = \sum_{d | n} \mu(d) \frac{n}{d}$$ Then you prove that this function $\mu(n)$ having the property that $\sum_{d | n} \mu(d) = \begin{cases}1 \text{ if } n = 1,\\ 0 \text{ otherwise } \end{cases}$ is $$\mu(n) = \begin{cases}\prod_{p | n} (-1) \text{ if } n \text{ is square-free },\\ 0 \text{ otherwise } \end{cases}$$ From which what you wrote follows