A right module is flat exactly when Tor one against every left module vanishes

If $N$ is flat, tensor a full projective resolution of $M$ with $N$. Exactness of $N{\otimes }_R-$ preserves the augmented resolution, so its positive homology, and in particular ${\mathrm{Tor}}_1(N,M)$ by The balanced Tor bifunctor, vanishes.

If $A\to B\to C\to 0$ is exact in left modules, Universal property of the tensor product for balanced maps into abelian groups identifies $N{\otimes }_{R}C$ with $(N{\otimes }_{R}B)\mathrm{/}\mathrm{im}(N{\otimes }_{R}A)$: balanced maps annihilating the image are precisely those that descend to $N\times C$. Thus $N{\otimes }_R-$ is right exact over the arbitrary ring $R$. Applied to a resolution $P_1\to P_0\to X\to 0$, this gives $H_0(N{\otimes }_R{P}_{\bullet })\cong N{\otimes }_{R}X$. This is natural under comparison maps because augmentations commute with them. By The balanced Tor bifunctor, it is the natural identification ${\mathrm{Tor}}_0^R(N,X)\cong N{\otimes }_{R}X$.

Conversely, for an inclusion $K\hookrightarrow P$ of left modules, apply The long exact Tor sequence in the left-module variable to $0\to K\to P\to P\mathrm{/}K\to 0$. Exactness identifies the kernel of $N\otimes K\to N\otimes P$ with the image of ${\mathrm{Tor}}_1(N,P\mathrm{/}K)$.

Universal vanishing gives injectivity for every such inclusion, and right exactness supplies the rest; thus $N{\otimes }_R-$ is exact.