Intuitive basis of Mobius inversion? If we're given $f(n)= \sum_{d|n}g\left(\frac{n}{d}\right),n \in \mathbb{N},$ then Mobius inversion gives
$$g(n)=\sum_{d|n}\mu \left( d\right) f \left( \frac{n}{d}\right).$$
Also, the generalised Mobius inversion formula states that the above is correct for $n \in \mathbb{R}$ if we change $d|n \to1\le d\le n$.
I'm familiar with the standard proof of the vanilla and generalised inversion formulae, but it relies on blind algebraic manipulation of Dirichlet convolution (and a generalised convolution between arithmetic and non-arithmetic functions), which I find unsatisfactory. 

Is there a more intuitive way of looking at the (generalised/ungeneralised) Mobius inversion formula, perhaps with knowledge garnered from more advanced theory?


The inversion formula can be discovered by working with many examples. Write out $f(1) = g(1)$, $f(2) = g(2) + g(1)$, $f(3) = g(3) + g(1)$, $f(4) = g(4) + g(2) + g(1)$, and so on, and then successively solve for $g(n)$ in terms of $f$-values:
$$
g(1) = f(1), g(2) = f(2) - f(1), g(3) = f(3) - f(1), g(4) = f(4) - f(2)
$$
and so on.  With enough data you discover $g(n)$ is a sum of $\pm f(n/d)$ for $d \mid n$ where $d$ is always squarefree. Understanding the pattern in the $\pm$ coefficients leads you eventually to the Moebius function.
The Moebius function is a very unnatural thing when you first see it. There is no way around that. By working with it enough, the function doesn't feel so strange anymore.
The most important property of the Moebius function is
that $\mu(1) = 1$ and $\sum_{d \mid n} \mu(d) = 0$ for $n > 1$. These properties in fact characterize it (only one sequence of numbers can fit these conditions).  And these are the properties you need to prove Moebius inversion.
There are efficient ways of explaining Moebius inversion using its role as coefficients in the Dirichlet series $1/\zeta(s) = \sum_{n \geq 1} \mu(n)/n^s$, but I don't know if you are familiar with Dirichlet series.
