The strategy of the proof will be: first reduce to the case where $\mathcal{L}$ and $\mathcal{N}$ are trivial invertible modules, then change our choices of local trivializations, and then finally use \'etale localization to reduce to the case of schemes\footnote{It is possible that a shorter proof can be given by immediately applying \'etale localization.}. \medskip\noindent First step. Let $q : T \to X$ be the morphism constructed in Lemma \ref{lemma-Gm-torsor}. We will use all properties stated in that lemma without further mention. In particular, it suffices to show that the cycles are equal after pulling back by $q$. Denote $s'$ and $t'$ the pullbacks of $s$ and $t$ to meromorphic sections of $q^*\mathcal{L}$ and $q^*\mathcal{N}$. Denote $Z'_i = q^{-1}(Z_i)$, denote $\xi'_i \in |Z'_i|$ the generic point, denote $B'_i = \mathcal{O}_{T, \xi'_i}^h$, denote $\mathcal{L}_{\xi'_i}$ and $\mathcal{N}_{\xi'_i}$ the pullbacks of $\mathcal{L}$ and $\mathcal{N}$ to $\Spec(B'_i)$. Recall that we have commutative diagrams $$ \xymatrix{ \Spec(B'_i) \ar[r]_-{c_{\xi'_i}} \ar[d] & T \ar[d]^q \\ \Spec(B_i) \ar[r]^-{c_{\xi_i}} & X } $$ see Decent Spaces, Remark \ref{decent-spaces-remark-functoriality-henselian-local-ring}. Denote $s'_i$ and $t'_i$ the pullbacks of $s_i$ and $t_i$ which are generators of $\mathcal{L}_{\xi'_i}$ and $\mathcal{N}_{\xi'_i}$. Then we have $$ s' = f'_i s'_i \quad\text{and}\quad t' = g'_i t'_i $$ where $f'_i$ and $g'_i$ are the images of $f_i, g_i$ under the map $Q(B_i) \to Q(B'_i)$ induced by $B_i \to B'_i$. By Algebra, Lemma \ref{algebra-lemma-pullback-module} we have $$ \text{ord}_{B_i}(f_i) = \text{ord}_{B'_i}(f'_i) \quad\text{and}\quad \text{ord}_{B_i}(g_i) = \text{ord}_{B'_i}(g'_i) $$ By Lemma \ref{lemma-prepare-flat-pullback-cap-c1} applied to $q : Z'_i \to Z_i$ we have $$ q^*\text{div}_{\mathcal{N}|_{Z_i}}(t_i|_{Z_i}) = \text{div}_{q^*\mathcal{N}|_{Z'_i}}(t'_i|_{Z'_i}) \quad\text{and}\quad q^*\text{div}_{\mathcal{L}|_{Z_i}}(s_i|_{Z_i}) = \text{div}_{q^*\mathcal{L}|_{Z'_i}}(s'_i|_{Z'_i}) $$ This already shows that the first cycle in the statement of the lemma pulls back to the corresponding cycle for $s', t', Z'_i, s'_i, t'_i$. To see the same is true for the second, note that by Chow Homology, Lemma \ref{chow-lemma-tame-symbol-formally-smooth} we have $$ \partial_{B_i}(f_i, g_i) \mapsto \partial_{B'_i}(f'_i, g'_i) \quad\text{via}\quad \kappa(\xi_i) \to \kappa(\xi'_i) $$ Hence the same lemma as before shows that $$ q^*\text{div}(\partial_{B_i}(f_i, g_i)) = \text{div}(\partial_{B'_i}(f'_i, g'_i)) $$ Since $q^*\mathcal{L} \cong \mathcal{O}_T$ we find that it suffices to prove the equality in case $\mathcal{L}$ is trivial. Exchanging the roles of $\mathcal{L}$ and $\mathcal{N}$ we see that we may similarly assume $\mathcal{N}$ is trivial. This finishes the proof of the first step. \medskip\noindent Second step. Assume $\mathcal{L} = \mathcal{O}_X$ and $\mathcal{N} = \mathcal{O}_X$. Denote $1$ the trivializing section of $\mathcal{L}$. Then $s_i = u \cdot 1$ for some unit $u \in B_i$. Let us examine what happens if we replace $s_i$ by $1$. Then $f_i$ gets replaced by $u f_i$. Thus the first part of the first expression of the lemma is unchanged and in the second part we add $$ \text{ord}_{B_i}(g_i)\text{div}(u|_{Z_i}) $$ where $u|_{Z_i}$ is the image of $u$ in the residue field by Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-divisor-meromorphic-well-defined} and in the second expression we add $$ \text{div}(\partial_{B_i}(u, g_i)) $$ by bi-linearity of the tame symbol. These terms agree by the property of the tame symbol given in Chow Homology, Equation (\ref{chow-item-normalization}). \medskip\noindent Let $Y \subset X$ be an integral closed subspace with $\dim_\delta(Y) = n - 2$. To show that the coefficients of $Y$ of the two cycles of the lemma is the same, we may do a replacement of $s_i$ by $1$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement of $t_i$ by $1$. Since there are only a finite number of $Z_i$ such that $Y \subset Z_i$ we may assume $s_i = 1$ and $t_i = 1$ for all these $Z_i$. \medskip\noindent Third step. Here we prove the coefficients of $Y$ in the cycles of the lemma agree for an integral closed subspace $Y$ with $\dim_\delta(Y) = n - 2$ such that moreover $\mathcal{L} = \mathcal{O}_X$ and $\mathcal{N} = \mathcal{O}_X$ and $s_i = 1$ and $t_i = 1$ for all $Z_i$ such that $Y \subset Z_i$. After replacing $X$ by a smaller open subspace we may in fact assume that $s_i$ and $t_i$ are equal to $1$ for all $i$. In this case the first cycle is zero. Our task is to show that the coefficient of $Y$ in the second cycle is zero as well. \medskip\noindent First, since $\mathcal{L} = \mathcal{O}_X$ and $\mathcal{N} = \mathcal{O}_X$ we may and do think of $s, t$ as rational functions $f, g$ on $X$. Since $s_i$ and $t_i$ are equal to $1$ we find that $f_i$, resp.\ $g_i$ is the image of $f$, resp.\ $g$ in $Q(B_i)$ for all $i$. Let $\zeta \in |Y|$ be the generic point. Choose an \'etale neighbourhood $$ (U, u) \longrightarrow (X, \zeta) $$ and denote $Y' = \overline{\{u\}} \subset U$. Since an \'etale morphism is flat, we can pullback $f$ and $g$ to regular meromorphic functions on $U$ which we will also denote $f$ and $g$. For every prime divisor $Y \subset Z \subset X$ the scheme $Z \times_X U$ is a union of prime divisors of $U$. Conversely, given a prime divisor $Y' \subset Z' \subset U$, there is a prime divisor $Y \subset Z \subset X$ such that $Z'$ is a component of $Z \times_X U$. Given such a pair $(Z, Z')$ the ring map $$ \mathcal{O}_{X, \xi}^h \to \mathcal{O}_{U, \xi'}^h $$ is \'etale (in fact it is finite \'etale). Hence we find that $$ \partial_{\mathcal{O}_{X, \xi}^h}(f, g) \mapsto \partial_{\mathcal{O}_{U, \xi'}^h}(f, g) \quad\text{via}\quad \kappa(\xi) \to \kappa(\xi') $$ by Chow Homology, Lemma \ref{chow-lemma-tame-symbol-formally-smooth}. Thus Lemma \ref{lemma-etale-pullback-principal-divisor} applies to show $$ (Z \times_X U \to Z)^*\text{div}_Z(\partial_{\mathcal{O}_{X, \xi}^h}(f, g)) = \sum\nolimits_{Z' \subset Z \times_X U} \text{div}_{Z'}(\partial_{\mathcal{O}_{U, \xi'}^h}(f, g)) $$ Since flat pullback commutes with pushforward along closed immersions (Lemma \ref{lemma-flat-pullback-proper-pushforward}) we see that it suffices to prove that the coefficient of $Y'$ in $$ \sum\nolimits_{Z' \subset U}  (Z' \to U)_*\text{div}_{Z'}(\partial_{\mathcal{O}_{U, \xi'}^h}(f, g)) $$ is zero. \medskip\noindent Let $A = \mathcal{O}_{U, u}$. Then $f, g \in Q(A)^*$. Thus we can write $f = a/b$ and $g = c/d$ with $a, b, c, d \in A$ nonzerodivisors. The coefficient of $Y'$ in the expression above is $$ \sum\nolimits_{\mathfrak q \subset A\text{ height }1} \text{ord}_{A/\mathfrak q}(\partial_{A_\mathfrak q}(f, g)) $$ By bilinearity of $\partial_A$ it suffices to prove $$ \sum\nolimits_{\mathfrak q \subset A\text{ height }1} \text{ord}_{A/\mathfrak q}(\partial_{A_\mathfrak q}(a, c)) $$ is zero and similarly for the other pairs $(a, d)$, $(b, c)$, and $(b, d)$. This is true by Chow Homology, Lemma \ref{chow-lemma-key-nonzerodivisors}.