The statement that $[\mathcal{F}]_r$ and $[\mathcal{G}]_s$ intersect properly is immediate. Since we are proving an equality of cycles we may work locally on $X$. (Observe that the formation of the intersection product of cycles, the formation of $\text{Tor}$-sheaves, and forming the cycle associated to a coherent sheaf, each commute with restriction to open subschemes.) Thus we may and do assume that $X$ is affine. \medskip\noindent Denote $$ RHS(\mathcal{F}, \mathcal{G}) = [\mathcal{F}]_r \cdot [\mathcal{G}]_s \quad\text{and}\quad LHS(\mathcal{F}, \mathcal{G}) = \sum (-1)^p [\text{Tor}_p^{\mathcal{O}_X}(\mathcal{F}, \mathcal{G})]_{r + s - \dim(X)} $$ Consider a short exact sequence $$ 0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0 $$ of coherent sheaves on $X$ with $\text{Supp}(\mathcal{F}_i) \subset \text{Supp}(\mathcal{F})$, then both $LHS(\mathcal{F}_i, \mathcal{G})$ and $RHS(\mathcal{F}_i, \mathcal{G})$ are defined for $i = 1, 2, 3$ and we have $$ RHS(\mathcal{F}_2, \mathcal{G}) = RHS(\mathcal{F}_1, \mathcal{G}) + RHS(\mathcal{F}_3, \mathcal{G}) $$ and similarly for LHS. Namely, the support condition guarantees that everything is defined, the short exact sequence and additivity of lengths gives $$ [\mathcal{F}_2]_r = [\mathcal{F}_1]_r  + [\mathcal{F}_3]_r $$ (Chow Homology, Lemma \ref{chow-lemma-additivity-sheaf-cycle}) which implies additivity for RHS. The long exact sequence of $\text{Tor}$s $$ \ldots \to \text{Tor}_1(\mathcal{F}_3, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_1, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_2, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_3, \mathcal{G}) \to 0 $$ and additivity of lengths as before implies additivity for LHS. \medskip\noindent By Algebra, Lemma \ref{algebra-lemma-filter-Noetherian-module} and the fact that $X$ is affine, we can find a filtration of $\mathcal{F}$ whose graded pieces are structure sheaves of closed subvarieties of $\text{Supp}(\mathcal{F})$. The additivity shown in the previous paragraph, implies that it suffices to prove $LHS = RHS$ with $\mathcal{F}$ replaced by $\mathcal{O}_V$ where $V \subset \text{Supp}(\mathcal{F})$. By symmetry we can do the same for $\mathcal{G}$. This reduces us to proving that $$ LHS(\mathcal{O}_V, \mathcal{O}_W) = RHS(\mathcal{O}_V, \mathcal{O}_W) $$ where $W \subset \text{Supp}(\mathcal{G})$ is a closed subvariety. If $\dim(V) = r$ and $\dim(W) = s$, then this equality is the {\bf definition} of $V \cdot W$. On the other hand, if $\dim(V) < r$ or $\dim(W) < s$, i.e., $[V]_r = 0$ or $[W]_s = 0$, then we have to prove that $RHS(\mathcal{O}_V, \mathcal{O}_W) = 0$ \footnote{The reader can see that this is not a triviality by taking $r = s = 1$ and $X$ a nonsingular surface and $V = W$ a closed point $x$ of $X$. In this case there are $3$ nonzero $\text{Tor}$s of lengths $1, 2, 1$ at $x$.}. \medskip\noindent Let $Z \subset V \cap W$ be an irreducible component of dimension $r + s - \dim(X)$. This is the maximal dimension of a component and it suffices to show that the coefficient of $Z$ in $RHS$ is zero. Let $\xi \in Z$ be the generic point. Write $A = \mathcal{O}_{X, \xi}$, $B = \mathcal{O}_{X \times X, \Delta(\xi)}$, and $C = \mathcal{O}_{V \times W, \Delta(\xi)}$. By Lemma \ref{lemma-tor-and-diagonal} we have $$ \text{coeff of }Z\text{ in } RHS(\mathcal{O}_V, \mathcal{O}_W) =  \sum (-1)^i \text{length}_B \text{Tor}_i^B(A, C) $$ Since $\dim(V) < r$ or $\dim(W) < s$ we have $\dim(V \times W) < r + s$ which implies $\dim(C) < \dim(X)$ (small detail omitted). Moreover, the kernel $I$ of $B \to A$ is generated by a regular sequence of length $\dim(X)$ (Lemma \ref{lemma-diagonal-regular-immersion}). Hence vanishing by Lemma \ref{lemma-one-ideal-ci} because the Hilbert function of $C$ with respect to $I$ has degree $\dim(C) < n$ by Algebra, Proposition \ref{algebra-proposition-dimension}.