Suppose E be a right R-module, and E is flat, then how can I prove that for any exact sequence of left R-modules $A\to B\to C$, the sequence $E\otimes A\to E\otimes B\to E\otimes C$ is exact? I am reaaly confused
E is flat means that Given any exact sequence $0\to A'\to A$, $E\otimes A'\to E\otimes A$ is injective (over here A' and A are just left R-modules)
Let $\alpha: A \to B$ and $\beta: B \to C$ be the maps. Then the sequence $A \to B \to C$ is exact (at B) if and only if $$ 0 \to \alpha (A) \to B \to B/\ker \beta \to 0$$ is exact.
Similarly, $E \otimes A \to E \otimes B \to E \otimes C$ is exact if and only if
$$0 \to (\mathbf{1} \otimes \alpha )(A) \to E \otimes B \to E \otimes B/ \ker \mathbf{1} \otimes \beta \to 0 $$ is exact. Observe that this sequence is obtained by tensoring the first short exact sequence by $E$.
Now tensoring is a right-exact functor. Flatness, in the definition given in your question, gives left-exact.