Tried to improve explanation

Thanks to Bogdan
https://stacks.math.columbia.edu/tag/0BVI#comment-6316

discriminant.tex

@@ -3257,7 +3257,7 @@ \section{Comparison with duality theory}
 \text{Trace}_f : f_*\mathcal{O}_Y \longrightarrow \mathcal{O}_X
 $$
 of Section \ref{section-discriminant}
-corresponds to a map $\mathcal{O}_Y \to f^!\mathcal{O}_X$.
+corresponds to a map $\mathcal{O}_Y \to f^!\mathcal{O}_X$ (see proof).
 Denote $\tau_{Y/X} \in H^0(Y, f^!\mathcal{O}_X)$ the image of $1$.
 Via the isomorphism $H^0(f^!\mathcal{O}_X) = \omega_{X/Y}$ of
 Lemma \ref{lemma-compare-dualizing}
@@ -3266,10 +3266,48 @@ \section{Comparison with duality theory}
 \end{lemma}
 
 \begin{proof}
-Unwinding all the definitions, this is immediate from the fact
-that if $A \to B$ is finite flat, then $\tau_{B/A} = \text{Trace}_{B/A}$
-(Lemma \ref{lemma-finite-flat-trace}) and the compatibility
-of traces with localizations (Lemma \ref{lemma-trace-base-change}).
+The functor $f^!$ is defined in 
+Duality for Schemes, Section \ref{duality-section-upper-shriek}.
+Since $f$ is finite (and hence proper), we see that $f^!$ is given by
+the right adjoint to pushforward for $f$. In
+Duality for Schemes, Section \ref{duality-section-duality-finite}
+we have made this adjoint explicit. In particular,
+the object $f^!\mathcal{O}_X$ consists of a single
+cohomology sheaf placed in degree $0$ and for this sheaf we have
+$$
+f_*f^!\mathcal{O}_X =
+\SheafHom_{\mathcal{O}_X}(f_*\mathcal{O}_Y, \mathcal{O}_X)
+$$
+To see this we use also that $f_*\mathcal{O}_Y$ is finite locally free
+as $f$ is a finite flat morphism of Noetherian schemes
+and hence all higher Ext sheaves are zero. Some details omitted.
+Thus finally
+$$
+\text{Trace}_f \in
+\Hom_{\mathcal{O}_X}(f_*\mathcal{O}_Y, \mathcal{O}_X) =
+\Gamma(X, f_*f^!\mathcal{O}_X) =
+\Gamma(Y, f^!\mathcal{O}_X)
+$$
+On the other hand, we have $f^!\mathcal{O}_X = \omega_{Y/X}$
+by the identification of Lemma \ref{lemma-compare-dualizing}.
+Thus we now have two elements, namely $\text{Trace}_f$
+and $\tau_{Y/X}$ from
+Remark \ref{remark-relative-dualizing-for-flat-quasi-finite} in
+$$
+\Gamma(Y, f^!\mathcal{O}_X) = \Gamma(Y, \omega_{Y/X})
+$$
+and the lemma says these elements are the same.
+
+\medskip\noindent
+Let $U = \Spec(A) \subset X$ be an affine open with inverse image
+$V = \Spec(B) \subset Y$. Since $f$ is finite, we see that
+$A \to B$ is finite and hence the $\omega_{Y/X}(V) = \Hom_A(B,A)$
+by construction and this isomorphism agrees with the identification
+of $f_*f^!\mathcal{O}_Y$ with
+$\SheafHom_{\mathcal{O}_X}(f_*\mathcal{O}_Y, \mathcal{O}_X)$ discussed
+above. Hence the agreement of $\text{Trace}_f$ and $\tau_{Y/X}$
+follows from the fact that $\tau_{B/A} = \text{Trace}_{B/A}$
+by Lemma \ref{lemma-finite-flat-trace}.
 \end{proof}