Let me just add a community wiki answer describing my understanding of Tate's paper Residues of differentials on curves.
The main theorem is as follows:
First, for any two subspaces $A,B\subseteq V$ of a $k$-vector space, we define a relation $A<B\Leftrightarrow (A+B)/B$ is finite dimensional, so this means $A$ is not much bigger than $B$. We also define an equivalence relation $A\sim B$ iff $A<B$ and $B<A$.
Let $K$ be a commutative $k$-algebra with unity,
$V$ a $K$-module, and $A$ a $k$-subspace of $V$ such that $fA< A$ for all $f\in K$. If we denote $E$ to be the $k$-subspace of $\mathrm{End}_k(V)$ that maps $A$ to $A$, then this just means the action of $K$ (by scalar multiplication) on $V$ factors through $E$. We define two further subspaces $E_1$ and $E_2$ of $E$, where $E_1:=\{\varphi\in \mathrm{End}_k(V): \varphi(V)< A\}$ and $E_2:=\{\varphi\in \mathrm{End}_k(V): \varphi(A)<0\}$. Note that $E=E_1+E_2$ since if we let $\pi:V\to A$ be the projection then we have $\pi\in E_1$ and $1-\pi\in E_2$.
Then there is a unique $k$-linear residue map res$_A^V:\Omega_{K/k}^1\to k$ such that for each pair of elements $f$ and $g$ in $K$ we have $$\mathrm {res}_A^V(fdg)=\mathrm{Tr}_V([f_1,g_1])$$ whenever $f_1,g_1\in E$ satisfies
$(1) f=f_1\mod E_2, \quad g=g_1\mod E_2$
$(2)$ Either $f_1\in E_1$ or $g_1\in E_1$
Such a pair $f_1,g_1$ always exists since $E=E_1+E_2$.
In the major application of this result, we have $K=k(X)$ the function field of a proper regular curve, $V=K_p$ the completion of $K$ at a closed point $p$ of $X$, and $A=\hat{O}_p$ the completed local ring at $p$. The spaces $E_1$ and $E_2$ are needed so that $\mathrm{Tr}_V([f_1,g_1])$ can be defined since it will turn out $[f_1,g_1]\in E_1\cap E_2$ (the reason it is in $E_2$ is beause $[f_1,g_1]=[f,g]\mod E_2$ and $[f,g]=0$ since they are elements of the commutative algebra $K$!) and therefore it is finitepotent (i.e. the image of a big enough power lies in a finite dimensional space) so we can define trace.
Where does the map $\mathrm{res}_A^V$ comes from? Recall that the Kahler differential is defined by $\Omega^1_{K/k}\cong (K\otimes_k K)/I$ via sending $f\otimes g\mapsto fdg$ and $I$ is essentially consists of the relations encoding the Leibninz's rule. So it suffices to construct a $k$-bilinear map $r:K\otimes K\to k$ and show that it factors through $I$. We simply define $e(f\otimes g)=\mathrm{Tr}_V([f_1,g_1])$. The fact that this factors through $I$ turns out to be the Jacobi identity for commutator.
To see why $\mathrm{res}_A^V$ is the actual residue. Let's take a $k$-rational point $p$ of $X$, so $A_p=k[[t]]$ and $K_p=k((t))$. Then to compute e.g. $\mathrm{res}(g^ndg)$, we choose $g_1=g\mod E_2$ and $g_1\in E_1$. Then for $n\ge 0$ we have $\mathrm{res}(g^n\otimes g)=\mathrm{Tr}([g_1^n,g_1])=0$ since $g_1$ obviously commute with a power of $g_1$. For $n\le -2$ and $g$ is invertible in $K$ (which is the case when $K$ is a field), we have $g^{-2-n}dg=-(g^{-1})^nd(g^{-1})$ so we can apply the preceding case to $g^{-1}$ (note that while $g$ is invertible $g_1\in E_1$ need not)
The finally most interesting case is to compute $\mathrm{res}(g^{-1}dg)$. Let $\pi$ be a projection of $V$ onto $A$, by definition we have $\mathrm{res}(fdg)=\mathrm{Tr}_V([\pi f,g])$. Note that $[\pi f,g]$ maps $V$ into $B:=A+gA$ and maps $C:=\{v\in B: fv\in A, fgv\in A\}$ to $0$ (because $fg=gf$). Hence $Tr_V([\pi f,g])=Tr_{B/C}([\pi f,g])$ by properties of trace. Applying this in the case of interest, we see that $$\mathrm{res}(g^{-1}dg)=\mathrm{Tr}(\pi-g\pi g^{-1})=\mathrm{Tr}_{(A+gA)/(A\cap gA)}(\pi)-\mathrm{Tr}_{(A+gA)/(A\cap gA)}(g\pi g^{-1})=\dim(A/(A\cap gA))-\dim(gA/(A\cap gA))$$ since $g\pi g^{-1}$ is a projection of $V$ onto $gA$. Hence if $gA\subseteq A$, then $\mathrm{res}(g^{-1}dg)=\dim(A/gA)$. In particular $\mathrm{res}(t^{-1}dt)=1$, which fits our expectation. More generally it is not hard to see that
$$\mathrm{res} (f dg) = \text{coefficient of } t^{-1}
\text{ in } f(t)g'(t)$$
I think this purely algebraic approach to residue is really beautiful and you actually see a concrete trace map being constructed (you probably also know that in the more general Serre duality situation the core of the argument is the construction of the trace map, but in that case it is much more abstract).
Finally, the biggest highlight of the paper is to prove that the sum of residues of a rational differential $\omega\in \Omega_{K/k}^1$ over a complete curve is zero, and this (magically!) boils down to the fact that $H^0(X,\mathcal O_X)$ is finite dimensional. I highly recommend anyone curious to read the paper to find out. In it Tate also proved the functoriality of $\mathrm{res}$ and the special case of Serre duality that $\Omega_{K/k}^1$ is a dualizing sheaf.
