If $\mathfrak q$ is a height $1$ prime of $A$ such that $a, b$ map to a unit of $A_\mathfrak q$, then $\partial_{A_\mathfrak q}(a, b) = 1$. Thus the sum is finite. In fact, if $V(ab) = \{\mathfrak m, \mathfrak q_1, \ldots, \mathfrak q_r\}$ then the sum is over $i = 1, \ldots, r$. For each $i$ we pick an extension $A_{\mathfrak q_i} \subset B_i$ as in Lemma \ref{lemma-prepare-tame-symbol} for $a, b$. By Lemma \ref{lemma-perpare-key} with $t = ab$ and the given list of primes we may assume we have a finite local extension $A \subset B$ with $B/A$ annihilated by a power of $ab$ and such that for each $i$ the $B_{\mathfrak q_i} \cong B_i$. Observe that if $\mathfrak q_{i, j}$ are the primes of $B$ lying over $\mathfrak q_i$ then we have $$ \text{ord}_{A/\mathfrak q_i}(\partial_{A_{\mathfrak q_i}}(a, b)) = \sum\nolimits_j \text{ord}_{B/\mathfrak q_{i, j}}(\partial_{B_{\mathfrak q_{i, j}}}(a, b)) $$ by Lemma \ref{lemma-norm-down-tame-symbol} and Algebra, Lemma \ref{algebra-lemma-finite-extension-dim-1}. Thus we may replace $A$ by $B$ and reduce to the case discussed in the next paragraph. \medskip\noindent Assume for each $i$ there is a nonzerodivisor $\pi_i \in A_{\mathfrak q_i}$ and units $u_i, v_i \in A_{\mathfrak q_i}$ such that for some integers $e_i, f_i \geq 0$ we have $$ a = u_i \pi_i^{e_i},\quad b = v_i \pi_i^{f_i} $$ in $A_{\mathfrak q_i}$. Setting $m_i = \text{length}_{A_{\mathfrak q_i}}(A_{\mathfrak q_i}/\pi_i)$ we have $\partial_{A_{\mathfrak q_i}}(a, b) = ((-1)^{e_if_i}u_i^{f_i}v_i^{-e_i})^{m_i}$ by definition. Since $a, b$ are nonzerodivisors the $(2, 1)$-periodic complex $(A/(ab), a, b)$ has vanishing cohomology. Denote $M_i$ the image of $A/(ab)$ in $A_{\mathfrak q_i}/(ab)$. Then we have a map $$ A/(ab) \longrightarrow \bigoplus M_i $$ whose kernel and cokernel are supported in $\{\mathfrak m\}$ and hence have finite length. Thus we see that $$ \sum e_A(M_i, a, b) = 0 $$ by Lemma \ref{lemma-compare-periodic-lengths}. Hence it suffices to show $e_A(M_i, a, b) = - \text{ord}_{A/\mathfrak q_i}(\partial_{A_{\mathfrak q_i}}(a, b))$. \medskip\noindent Let us prove this first, in case $\pi_i, u_i, v_i$ are the images of elements $\pi_i, u_i, v_i \in A$ (using the same symbols should not cause any confusion). In this case we get \begin{align*} e_A(M_i, a, b) & = e_A(M_i, u_i\pi_i^{e_i}, v_i\pi_i^{f_i}) \\ & = e_A(M_i, \pi_i^{e_i}, \pi_i^{f_i}) - e_A(\pi_i^{e_i}M_i, 0, u_i) + e_A(\pi_i^{f_i}M_i, 0, v_i) \\ & = 0 - f_im_i\text{ord}_{A/\mathfrak q_i}(u_i) + e_im_i\text{ord}_{A/\mathfrak q_i}(v_i) \\ & = -m_i\text{ord}_{A/\mathfrak q_i}(u_i^{f_i}v_i^{-e_i}) = -\text{ord}_{A/\mathfrak q_i}(\partial_{A_{\mathfrak q_i}}(a, b)) \end{align*} The second equality holds by Lemma \ref{lemma-multiply-period-length}. Observe that $M_i \subset (M_i)_{\mathfrak q_i} = A_{\mathfrak q_i}/(\pi_i^{e_i + f_i})$ and $(\pi_i^{e_i}M_i)_{\mathfrak q_i} \cong A_{\mathfrak q_i}/\pi_i^{f_i}$ and $(\pi_i^{f_i}M_i)_{\mathfrak q_i} \cong A_{\mathfrak q_i}/\pi_i^{e_i}$. The $0$ in the third equality comes from Lemma \ref{lemma-powers-period-length-zero} and the other two terms come from Lemma \ref{lemma-length-multiplication}. The last two equalities follow from multiplicativity of the order function and from the definition of our tame symbol. \medskip\noindent In general, we may first choose $c \in A$, $c \not \in \mathfrak q_i$ such that $c\pi_i \in A$. After replacing $\pi_i$ by $c\pi_i$ and $u_i$ by $c^{-e_i}u_i$ and $v_i$ by $c^{-f_i}v_i$ we may and do assume $\pi_i$ is in $A$. Next, choose an $c \in A$, $c \not \in \mathfrak q_i$ with $cu_i, cv_i \in A$. Then we observe that $$ e_A(M_i, ca, cb) = e_A(M_i, a, b) - e_A(aM_i, 0, c) + e_A(bM_i, 0, c) $$ by Lemma \ref{lemma-length-multiplication}. On the other hand, we have $$ \partial_{A_{\mathfrak q_i}}(ca, cb) = c^{m_i(f_i - e_i)}\partial_{A_{\mathfrak q_i}}(a, b) $$ in $\kappa(\mathfrak q_i)^*$ because $c$ is a unit in $A_{\mathfrak q_i}$. The arguments in the previous paragraph show that $e_A(M_i, ca, cb) = - \text{ord}_{A/\mathfrak q_i}(\partial_{A_{\mathfrak q_i}}(ca, cb))$. Thus it suffices to prove $$ e_A(aM_i, 0, c) = \text{ord}_{A/\mathfrak q_i}(c^{m_if_i}) \quad\text{and}\quad e_A(bM_i, 0, c) = \text{ord}_{A/\mathfrak q_i}(c^{m_ie_i}) $$ and this follows from Lemma \ref{lemma-length-multiplication} by the description (see above) of what happens when we localize at $\mathfrak q_i$.