I came to this question hoping to learn more, but I can record what I've learned. I find the Cartier operator much more natural in a larger context. This is a combination of material from [Brion and Kumar's book][1], Section I.3, and things I saw asserted in papers and worked out for myself. Of course, there is a risk that any of this is wrong.$\def\cF{\mathcal{F}}$ $\def\cC{\mathcal{C}}$

Notation: Let $k$ be a perfect field of characteristic $p$ and $A$ a $k$-algebra. Let $\Omega^r$ be the $r$-th wedge power of the Kahler differentials of $A/k$. Let $Z^r = \mathrm{Ker}(d: \Omega^r \to \Omega^{r+1})$ (the de Rham co-cycles) and $B^r = \mathrm{Im}(d: \Omega^{r-1} \to \Omega^r)$ (the de Rham co-boundaries). Let $H^r = Z^r/B^r$. Let $A^p$ denote the ring of $p$-th powers in $A$. Note that $d$ is a map of $A^p$ modules and hence $Z^r$, $B^r$ and $H^r$ are all $A^p$ modules.

The inverse of the Cartier operator is more basic than the Cartier operator. 

<b>Theorem</b> There is a unique map $\cF: \Omega^r \to H^r$ such that

 - $\cF(f) = f^p$ for $f \in \Omega^0 = A$.

 - $\cF(\alpha+\beta) = \cF(\alpha)+\cF(\beta)$.

 - $\cF(d u) = u^{p-1} du$ for $u \in A$.

 - $\cF(\alpha \wedge \beta) = \cF(\alpha) \wedge \cF(\beta)$.

The only hard part to check is that $(u+v)^{p-1} (du+dv)$ is equivalent to $u^{p-1} du+ v^{p-1} dv$ modulo $B^1$. To this end, note that 
$$(u+v)^{p-1} (du+dv) - u^{p-1} du - v^{p-1} dv = d \sum_{k=1}^{p-1} \frac{1}{p} \binom{p}{k} d(u^k v^{p-k}). \quad (\ast)$$
Here $\frac{1}{p} \binom{p}{k}$ must be interpreted as an integer which we reduce modulo $p$ to make an element of $k$. (Morally, equation $(\ast)$ comes from the relation $d(u+v)^p = \sum_{k=0}^{p} \binom{p}{k} d(u^k v^{p-k})$. We can't deduce $(\ast)$ directly from this because we can't divide by $p$, but $(\ast)$ is still true in characteristic $p$.)

If $A$ has a sufficiently nice deformation $\tilde{A}$ over the ring of Witt vectors $W(k)$, and $F: \tilde{A} \to \tilde{A}$ is a lift of Frobenius, then $\cF$ for $\omega \in \Omega^r$ is given by reduction modulo $p$ of $\frac{1}{p^r} (F^{\ast})^r \omega$. I don't know exactly what "sufficiently nice" means, but $\tilde{A}$ smooth over $W(k)$ is certainly enough. This is my favorite way to think of $\cF$.

$\cF$ respects localization, and thus makes sense on schemes. It also respects etale extension and completion, for those who prefer other topologies.

We now have [Theorem I.3 in Brion and Kumar][2] (presumably not original, but they don't cite it): If $A$ is regular, than $\cF$ is an isomorphism $\Omega^{\bullet} \to H^{\bullet}$. Proof sketch: The claim is that a map of finitely generated $A^p$ modules is an isomorphism; this can be checked on completions. So we are reduced to proving the claim in a power series algebra. This is a combinatorial exercise, very similar to the proof of the Poincare lemma for formal power series in characteristic zero. 

Note that this is interesting even for $r=0$: It says that $df=0$ for $f \in A$ if and only if $f$ is a $p$-th power.

For $A$ regular, the Cartier operator $\cC: H^{\bullet} \to \Omega^{\bullet}$ is the inverse to $\cF$.

In particular, if $A$ regular of dimension $1$, then we can take the composition $\Omega^1 \to H^1 \to \Omega^1$ to get the definition you gave.

$\cC$ on a curve is analogous to residue in several senses: 

 - It is the composition of $\Omega^1 \to H^1$ and an isomorphism between $H^1$ and an explicit free $H^0$-module of rank $1$. In the case of residue, working in the Laurent series ring $k((x))$, we have $H^0 = k$ and we have a canonical isomorphism $H^1 \cong k$; the residue map is $\Omega^1 \to H^1 \to k$. In the Cartier case, $H^0 = A^p$ and $H^1$ is a free $A^p$-module of rank $1$. It doesn't have a canonical generator, but what you can do canonically is use $\cC$ to turn $A^p$ into $A$ and $H^1$ into $\Omega^1$.

 - We have $\cC(df) = 0$ and $\cC(du/u) = du/u$, for $u$ a unit of $A$.

 - Concretely, $\mathrm{res}(\cC(\omega)) = \mathrm{res}(\omega)^{1/p}$.

  [1]: http://books.google.com/books/about/Frobenius_Splitting_Methods_in_Geometry.html?id=i22xdTWr2XwC
  [2]: http://books.google.com/books?id=i22xdTWr2XwC&lpg=PP1&pg=PA24#v=onepage&q&f=false