Exact sequence with flat module tensored by module stays exact The following theorem is given in Liu's book Algebraic Geometry and Arithmetic Curves, proposition 1.2.6:
Let $A$ be a ring. Let $0\to M^\prime\to M\to M^{\prime\prime}\to 0$ be an exact sequence of $A$-modules. Let us suppose that $M^{\prime\prime}$ is flat. Then the sequence $0\to M^\prime \otimes N\to M\otimes N\to M^{\prime\prime} \otimes N\to 0$ is also exact for any $A$-module $N$.
Liu gives proof in http://176.58.104.245/ALGANT/TONG/Liu-1-4.pdf page 9.
I have problems to understand the proof. The mistake is easy to fix, Proposition 2.12 should be 1.12. But what does the "Let us write $N$ as the quotient of a free $A$-module $L$ by a submodule $K$" means? Is it $N=L/K$? And how one can do the diagram chasing part?
 A: Consider the diagram of the proof:
$$\require{AMScd}
\begin{CD}
{} @. M'\otimes K @>f>> M\otimes K @>g>> M''\otimes K \\
@. @VVhV @VViV @VV{\alpha}V \\
0 @>>> M'\otimes L @>j>> M\otimes L @>k>> M''\otimes L \\
@. @VV{\beta}V @VVlV \\
{} @. M'\otimes N @>m>> M\otimes N \\
@. @VVV @VVV \\
{} @. 0 @. 0
\end{CD}
$$
Let $x\in\ker m$; then $x=\beta(y)$ and $lj(y)=0$, so $j(y)=i(z)$.
Thus $\alpha g(z)=ki(z)=kj(y)=0$, so $g(z)=0$ and therefore $z=f(u)$.
Since $jh(u)=if(u)=i(z)=j(y)$, we have $y-h(u)\in\ker j=\{0\}$, so $y=h(u)$ and finally $x=\beta(y)=\beta h(u)=0$.
A: Yes, there is a surjective homomorphism of $A$-modules $L\to N$ and denote its kernel by $K$. Then, by the fundamental isomorphism theorem we have $N\simeq L/K$.
The diagram chasing isn't that hard: let $x\in M\otimes N$ which is sent to $0$; since $\beta$ is surjective there is $y\in M'\otimes L$ such that $x=\beta(y)$. Now let $f=(M\otimes L\to M\otimes N)$, and $g=(M'\otimes L\to M\otimes L)$. From the commutativity of the diagram we have $(f\circ g)(y)=0$, so $g(y)\in\ker f$. But $\ker f=\operatorname{im}h$, where $h=(M\otimes K\to M\otimes L)$, so $g(y)=h(z)$, $z\in M\otimes K$. Let $u=(M\otimes L\to M''\otimes L)$ and $v=(M\otimes K\to M''\otimes K)$. We have $(\alpha\circ v)(z)=(u\circ h)(z)=u(h(z))=u(g(y))=0$ since $\operatorname{im}g=\ker u$. So, $\alpha(v(z))=0$ and since $\alpha$ is injective we get $v(z)=0$, that is, $z\in\ker v$. Then there is $z'\in M'\otimes K$ such that $z=w(z')$, where $w=(M'\otimes K\to M\otimes K)$. We have $(h\circ w)(z')=h(w(z'))=h(z)=g(y)$. On the other side, $h\circ w=g\circ h'$, where $h'=(M'\otimes K\to M'\otimes L)$, so $g(y)=g(h'(z'))$ and since $g$ is injective we get $y=h'(z')$. The final step is to recall that $x=\beta(y)=(\beta\circ h')(z')=0$ for $\operatorname{im}h'=\ker\beta$.
Remark. Instead of the above diagram chasing you can use the Snake lemma (which is the right tool for such things). All your data fit well into the lemma, the last kernel is $0$ (as being $\ker\alpha$), while the first two cokernels are $M'\otimes N$, respectively $M\otimes N$ (because the vertical maps are surjective).
A: $$0=Tor^1(M^{\prime\prime},N)\to  M^\prime \otimes N\to M\otimes N\to M^{\prime\prime} \otimes N\to 0$$ 
$0=Tor^1(M^{\prime\prime},N)$, since $M^{\prime\prime}$ is flat.
