Let $R$ be a ring with unity, $A$ a right $R$-module, $B$ a left $R$-module,
and $\cdots\rightarrow P_2\rightarrow P_1\rightarrow P_0\xrightarrow{f} A\rightarrow 0$ a projective resolution of $A$.
Consider $K=\ker{f}$ and the exact sequence $K\rightarrow P_0\rightarrow A\rightarrow 0$ and applying tensor product with $B$, $K\otimes B\xrightarrow{u} P_0\otimes B\rightarrow A\otimes B\rightarrow 0$(exact since tensor product is right exact.)
Then $Tor_{1}^{R}(A,B)=\ker{u}$.
But by definition, from $P_2\otimes B\xrightarrow{r} P_1\otimes B\xrightarrow{s} P_0\otimes B\rightarrow A\otimes B\rightarrow 0$, $Tor_{1}^{R}(A,B)=\ker{s}/\mathrm{Im}\ r$.
The equation, $Tor_{1}^{R}(A,B)=\ker{u}$ comes from the canonical isomorphism, $\ker{s}/\mathrm{Im}\ r\cong\ker{u}$?