Let $M$ be a flat $R$-module ($R$ is commutative). Then, $M$ is faithfully flat ($- \otimes_R M$ sends non-exact sequences to non-exact sequences) iff for any module homomorphism $f : N \rightarrow P$, $f \otimes id_M = 0$ implies $f = 0$.
The proof of $\rightarrow$ is : $0 \rightarrow ker(f) \rightarrow N \rightarrow P$ is exact iff the tensor product with $M$ is exact. That's the entire proof and (Question 1) I don't understand how it proves anything, it just restates the definition of faithful flatness.
Proof of $\leftarrow$ :
Let $0 \rightarrow A \rightarrow^f B \rightarrow^g C \rightarrow 0$ be a complex such that $0 \rightarrow A \otimes M \rightarrow B \otimes M \rightarrow C \otimes M \rightarrow 0$ is exact. Let $x \in ker(g)$ and we want to show that it's in $range(f)$. Define the map $a : R \rightarrow B / range(f)$ as $r \rightarrow rx$. Tensor product with $M$ to get $M \otimes_R R \rightarrow M \otimes_R (B / range(f))$ .
The next part is where I don't understand: It just says "By the exactness of the complex $- \otimes_R M$ we see that $a \otimes id_M$ is zero". I don't understand this. $rx$ is always in $ker(g)$, so $1 \otimes rx$ is always in $M \otimes ker(g) = ker(g \otimes id_M) = range(f \otimes id_M)$. (Question 2) But why is it zero in $M \otimes (B / range(f))$?
Question 1: The answer is that faithfully flat can be phrased as follows: if $0\to A\otimes_R M\to B\otimes_R M\to C\otimes_R M\to 0$ is exact, then $0\to A \to B \to C\to 0$ is exact. Or equivalently, the sequence you get from applying $-\otimes_R M$ is exact iff the original sequence was exact. This is why the adjective "faithfully" is added in front of flat.
Then Question 1 should come as no surprise: the sequence $0\to ker(f)\otimes_R M\to N\otimes_R M\stackrel{f\otimes_R id_M}{\to} P\otimes_R M$ has $f\otimes_R id_M$ equal to zero. So, $ker(f)\otimes_R M$ is zero. So $ker(f)\otimes_R M$ is all of $N\otimes_R M$. So $0\to ker(f)\to N\to 0$ is exact and this means $ker(f)=N$ aka $f=0$.
Question 2: The step in your last line of reasoning to finish it off is $rx\otimes 1$ is that you need to rewrite $M\otimes B/range(f)$. The sequence $0\to range(f)\to B\to B/range(f)\to 0$ tensored by $M$ implies $M\otimes B/range(f)= \frac{B\otimes M}{range(f)\otimes M}$.
If the preceding paragraph does not satisfy you, I wrote out extra details.
This answer to your second question is more or less due to the same reasoning as in Question 1. To show that the original complex is exact, you need to show if $x\in ker(g)$ then $x\in range(f)$. This is the same as showing $a:R\to B/range(f)$ is zero. Use the hypothesis to reduce to showing $a\otimes_R id_M$ is zero. The complex $0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$ is exact. I know $a\otimes_R id_M(r)=a\otimes_R id_M(r\otimes_R 1)=rx\otimes_R 1$. now apply $g\otimes_R id_M$ to this element to get $g\otimes_R id_M(rx\otimes_R 1)=0$ inside of $C\otimes M$. This implies $rx\otimes_R 1$ is in the image of $A\otimes M\to B\otimes M$. Taking $r=1$ we know $x\otimes 1$ is in the image. In particular, $x\otimes 1$ is in $range(f)\otimes M$. So $a\otimes_R id_M$ is zero. So $a=0$ and $x\in range(f)$ as desired.