The universal coefficient theorem for homology over a PID

The cycle and boundary modules are free, hence flat. Since $B_{n-1}C$ is flat, tensoring $0\to Z_{n}C\to C_n\to B_{n-1}C\to 0$ remains exact. The differential on $C_n\otimes G$ factors as the resulting surjection $C_n\otimes G\twoheadrightarrow B_{n-1}C\otimes G$ followed by the map $B_{n-1}C\otimes G\to Z_{n-1}C\otimes G\hookrightarrow C_{n-1}\otimes G$. The last inclusion stays injective because $B_{n-2}C$ is flat.

Tensoring $B_{n}C\to Z_{n}C\to H_n(C)\to 0$ gives $H_n(C)\otimes G=\mathrm{coker}(B_{n}C\otimes G\to Z_{n}C\otimes G)$. By step 1.1, $Z_{n}C\otimes G$ embeds in $C_n\otimes G$, and the image of $B_{n}C\otimes G$ there is exactly the boundary group. Thus the edge map $[z]\otimes g\mapsto [z\otimes g]$ injects $H_n(C)\otimes G$ into $H_n(C\otimes G)$.

The remaining quotient maps by $d_n\otimes 1$ onto $\ker (B_{n-1}C\otimes G\to Z_{n-1}C\otimes G)$. The free presentation $0\to B_{n-1}C\to Z_{n-1}C\to H_{n-1}C\to 0$ identifies this kernel with ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$. All maps come from the natural cycle, boundary and tensor maps, so the sequence is natural in $C$ and $G$.