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, 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.
Theorem 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.3.4 in Brion and Kumar (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, $\mathrm{res}(\cC(\omega)) = \mathrm{res}(\omega)^{1/p}$.
$\{\sigma = 0\}$. $\endgroup$