$\varphi(n)$ counts the number of integers $k$ between $1$ and $n$ that are relatively prime to $n$.
The term $d=n$ gives us our starting value of $n\cdot\mu(\frac{n}{n})=n\cdot 1=n$, i.e. all of the integers $1\leq k\leq n$.
For each prime $p$ dividing $n$, the term $d=\frac{n}{p}$ throws out the multiples of $p$ from among the numbers between $1$ and $n$; specifically, there are $\frac{n}{p}$ multiples of $p$ between $1$ and $n$, and the $d=\frac{n}{p}$ term in the sum contributes $\frac{n}{p}\cdot\mu(p)=-\frac{n}{p}$.
Unfortunately, the result is still not quite right - for any distinct prime divisors $p_1$ and $p_2$ of $n$, we counted as if we had to remove multiples of $p_1p_2$ twice, when of course we only remove them once. The terms $d=\frac{n}{p_1p_2}$ correct for this, because they contribute $\frac{n}{p_1p_2}\cdot \mu(p_1p_2)=\frac{n}{p_1p_2}$, the number of multiples of $p_1p_2$ between $1$ and $n$.
But then we must account for the numbers between $1$ and $n$ that we originally triple-counted, and which our correction above over corrects for, etc. - the alternating nature of the Möbius function is what makes it do this the way we need.