Proof.
We are going to deduce this from the case of schemes by étale localization. Let $Z \subset Y$ be an integral closed subspace of $\delta $-dimension $n - 1$. We want to show that the coefficient of $[Z]$ in $p_*\text{div}(f)$ and $\text{div}(g)$ are equal. Apply Spaces over Fields, Lemma 72.3.2 to the morphism $p : X \to Y$ and the generic point $\xi \in |Z|$. We find that we may replace $Y$ by an open subspace containing $\xi $ and assume that $p : X \to Y$ is finite. Pick an étale neighbourhood $(V, v) \to (Y, \xi )$ where $V$ is an affine scheme. By Lemma 82.10.3 it suffices to prove the equality of cycles after pulling back to $V$. Set $U = V \times _ Y X$ and consider the commutative diagram
\[ \xymatrix{ U \ar[r]_ a \ar[d]_{p'} & X \ar[d]^ p \\ V \ar[r]^ b & Y } \]
Let $V_ j \subset V$, $j = 1, \ldots , m$ be the irreducible components of $V$. For each $i$, let $U_{j, i}$, $i = 1, \ldots , n_ j$ be the irreducible components of $U$ dominating $V_ j$. Denote $p'_{j, i} : U_{j, i} \to V_ j$ the restriction of $p' : U \to V$. By the case of schemes (Chow Homology, Lemma 42.18.1) we see that
\[ p'_{j, i, *}\text{div}_{U_{j, i}}(f_{j, i}) = \text{div}_{V_ j}(g_{j, i}) \]
where $f_{j, i}$ is the restriction of $f$ to $U_{j, i}$ and $g_{j, i}$ is the norm of $f_{j, i}$ along the finite extension $R(U_{j, i})/R(V_ j)$. We have
\begin{align*} b^* p_*\text{div}_ X(f) & = p'_* a^* \text{div}_ X(f) \\ & = p'_*\left(\sum \nolimits _{j, i} (U_{j, i} \to U)_*\text{div}_{U_{j, i}}(f_{j, i})\right) \\ & = \sum \nolimits _{j, i} (V_ j \to V)_*p'_{j, i, *}\text{div}_{U_{j, i}}(f_{j, i}) \\ & = \sum \nolimits _ j (V_ j \to V)_*\left(\sum \nolimits _ i \text{div}_{V_ j}(g_{j, i})\right) \\ & = \sum \nolimits _ j (V_ j \to V)_*\text{div}_{V_ j}(\prod \nolimits _ i g_{j, i}) \end{align*}
by Lemmas 82.11.1, 82.13.2, and 82.8.2. To finish the proof, using Lemma 82.13.2 again, it suffices to show that
\[ g \circ b|_{V_ j} = \prod \nolimits _ i g_{j, i} \]
as elements of the function field of $V_ j$. In terms of fields this is the following statement: let $L/K$ be a finite extension. Let $M/K$ be a finite separable extension. Write $M \otimes _ K L = \prod M_ i$. Then for $t \in L$ with images $t_ i \in M_ i$ the image of $\text{Norm}_{L/K}(t)$ in $M$ is $\prod \text{Norm}_{M_ i/M}(t_ i)$. We omit the proof.
$\square$
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