I have problems working with the concept of flat modules. I am studying it with Atiyah-McDonald's , so I know the characterization of them as
Given $A$ a conmutative unity ring, an $A$-module $N$ being flat is equivalent to
(1) Any exact sequence $0 \rightarrow M'' \rightarrow M \rightarrow M' \rightarrow 0$ satisfies that its tensored sequence $0 \rightarrow M'' \otimes N \rightarrow M \otimes N \rightarrow M' \otimes N \rightarrow 0$ is exact.
(2) For every $A$-modules $M, M'$, if $f: M' \rightarrow M$ is injective then $f \otimes 1 : M' \otimes N \rightarrow M \otimes N$ is also injective.
For example, when I try to prove that given $A$-modules $M$ and $K \subset M$, if $M$ and $M/K$ are flat then $K$ is flat I am getting stuck. My first idea consists on using (1), taking an exact sequence $0 \rightarrow P'' \rightarrow P \rightarrow P' \rightarrow 0$, then $0 \rightarrow P'' \otimes M \rightarrow P \otimes M \rightarrow P' \otimes M \rightarrow 0$ and $0 \rightarrow P'' \otimes M/K \rightarrow P \otimes M/K \rightarrow P' \otimes M/K \rightarrow 0$ are exact and my intuition says that here should be a way to apply the Snake lemma, but I do not get it. Also, trying it with (2) it seems confusing, so any possible help would be appreciated.
EDIT
In order to prove the claim, I have considered the diagram
\begin{array}{c} & & 0 \rightarrow & P'' \otimes M & \xrightarrow{i \otimes Id_M} & P \otimes M & \xrightarrow{p \otimes Id_M} & P' \otimes M & \rightarrow 0 \\ & & & \downarrow_{Id_{P''} \otimes \pi} & & \downarrow_{Id_P \otimes \pi} & & \downarrow_{Id_{P'} \otimes \pi} & & \\ & & 0 \rightarrow & P'' \otimes M/K & \xrightarrow{i \otimes Id_{M/K}} & P \otimes M/K & \xrightarrow{p \otimes Id_{M/K}} & P' \otimes M/K & \rightarrow 0 \end{array}
where $\pi$ is the quotient morphism. Applying on it the Snake Lemma, the following sequence is exact: $$0 \rightarrow \ker Id_{P''} \otimes \pi \rightarrow \ker Id_{P} \otimes \pi \rightarrow \ker Id_{P'} \otimes \pi \rightarrow \operatorname{coker} Id_{P''} \otimes \pi \rightarrow \\ \rightarrow\operatorname{coker} Id_{P} \otimes \pi \rightarrow \operatorname{coker} Id_{P'} \otimes \pi \rightarrow 0 $$
If I could prove that $\ker Id_p \otimes \pi = P \otimes K$ (then similarly for $P', P''$) and that $\operatorname{coker} Id_P \otimes \pi = 0$, then it would be finished, but I am struggling with it, I do not have ideas since tensorial product is new and confusing for me yet, so any possible help or advice without using Homological algebra or more sophisticated concepts would be very apreciated.