Every $A$-mod is flat iff every mod-$A$ is flat. $\require{AMScd}$ Suppose $A$ is a ring, and we have a SES of left $A$-modules $$0\longrightarrow M'\mathop\longrightarrow\limits^f M \mathop\longrightarrow \limits^g M''\longrightarrow 0$$ where $M''$ is flat. Now let $N$ be any right $A$-module. Show that $$0\longrightarrow N\otimes M'\longrightarrow N \otimes M \longrightarrow N\otimes  M''\longrightarrow 0$$ is exact. Still assuming $M''$ is flat, show also that if $M'$ is flat, then so is $M$, and conversely. 
I am sure the second claim follows easily from the nine lemma, but I am stuck with the first one. The hint is to consider a SES $0 \longrightarrow K\longrightarrow F \longrightarrow N \longrightarrow 0$ where $F$ is free and analyze the resulting diagram obtained from tensoring term by term. One gets the following  $$\begin{CD} {}&{}&0&{}&0&{}&0\\
 {}&@VV(*)V@VV(*)V@VVV
  \\
0@>(?)>>
K\otimes M^\prime
  @>>>
  K\otimes M
  @>>> 
  K\otimes M^{\prime\prime}
  @>>> 0
  \\
{}& @VVV @VVV @VVV
  \\
0 @>>>
F\otimes M^\prime
  @>>>
 F\otimes M 
  @>>>
  F\otimes M^{\prime\prime}
  @>>>
  0
  \\
{}&@VVV @VVV @VVV
  \\
0 @>\rm prove>>
N\otimes M^\prime
  @>>>
  N\otimes M 
  @>>> 
 N\otimes M^{\prime\prime}
  @>>>
  0
  \\
{}& @VVV @VVV @VVV
  \\
{}&{}&0&{}&0&{}&0
\end{CD}$$
The second row is exact since $F$ is free hence flat, the third column is exact since $M''$ is flat. I want to prove the last row is exact, i.e. that $1_N\otimes f$ is injective, but I am not sure how to proceed. I tried some diagram chasing, but no luck. I am not sure $(?)$ the first row is exact i.e. that $1_K\otimes f$ is injective, since submodules of flat modules needn't be flat. Even assuming $(*)$ $M',M$ are both flat I do not get a diagram that allows the application of the nine lemma. At any rate, I don't want to make this assumption.
 A: Try using the snake lemma.
The second row is short exact, and the first row is almost short exact; you can make it short exact on the nose just by replacing $K \otimes M'$ with its image in $K \otimes M$.  (Note that by short exactness of the second row, the map from
$K \otimes M'$ to $F \otimes M'$ factors through this image.)
Now you have a map between two short exact sequences, and the snake lemma gives
a six term exact sequence.  The first two terms will be irrelevant, but the rest of them will give what you want.
[One way to think of this approach is that your (?) , while it generally has a negative answer --- you are right that submodules of flat modules need not be flat, and 
it's not hard to find examples where this map won't be injective --- this doesn't matter for the problem at hand!]
[And a general philosophical remark: when you have a diagram like the one you wrote, and you are trying to show that first arrow in a putative s.e.s. is injective, the snake lemma is a natural thing to use, so you want to try to set things up so that you putative s.e.s. is the last three terms of a snake lemma 6-term exact sequence.  This is made systematic by the theory of derived functors, and the theory of Tor is underlying this question.  More precisely, this question is a special case of the general fact that Tor${}^R_i(M,N)$ can be computed using a flat resolution of either $M$ or $N$.]
A: I will add the solution here. I am hoping all is correct. We have the diagram (I've removed some irrelevant arrows) $$\begin{CD} {}&{}&{}&{}&{}&{}&0\\
 {}&{}&{}&{}&{}&@VVV
  \\
&{}&
K\otimes M^\prime
  @>>>
  K\otimes M
  @>>> 
  K\otimes M^{\prime\prime}
  @>>> 0
  \\
{}& @VV\iota \otimes 1 V @VVV @VVV
  \\
0 @>>>
F\otimes M^\prime
  @>>>
 F\otimes M 
  @>>>
  F\otimes M^{\prime\prime}
  \\
{}&@VVV @VVV @VVV
  \\
&{}&
N\otimes M^\prime
  @>>>
  N\otimes M 
  @>>> 
 N\otimes M^{\prime\prime}
  \\
{}& @VVV @VVV @VVV
  \\
{}&{}&0&{}&0&{}&0
\end{CD}$$
Since the columns are exact, the various $\eta\otimes 1:F\otimes M'\longrightarrow N\otimes M'$ are cokernels for various $\iota\otimes 1$, and the uppermost $0$ is the kernel of the third $\iota\otimes 1$. Thus the snake lemma gives a connecting morphism $\partial$ making the sequence $0\mathop\longrightarrow\limits^{\partial}N\otimes M'\longrightarrow N\otimes M$ exact, as we wanted.
