It is true that if $R\to S$ is flat and $M$ is a flat $S$-module, then $M$ is flat as an $R$-module, but it is not true that if $R\to S$ is flat and $M$ is a faithfully flat $S$-module, then $M$ is faithfully flat as an $R$-module.
First claim: Let $M$ be a flat $S$-module, and let $$N'\to N\to N''$$ be an exact sequence of $R$-modules. We want to show that $$N'\otimes_R M\to N\otimes_R M\to N''\otimes M$$ is exact.
Since $R\to S$ is flat, we have that $$N'\otimes_R S\to N\otimes_R S\to N''\otimes_R S$$ is exact. Since $M$ is flat over $S,$ the sequence $$N'\otimes_R S\otimes_S M\to N\otimes_R S\otimes_S M\to N''\otimes_R S\otimes_S M$$ is exact as well. However, for any $R$-module $N,$ we have $N\otimes_R S\otimes_S M\cong N\otimes_R M,$ so $$N'\otimes_R M\to N\otimes_R M\to N''\otimes M$$ is exact, as desired.
Second claim: Now, suppose that $M$ is faithfully flat over $S.$ We already know that $M$ is flat over $R.$ Suppose that $N$ is an $R$-module such that $N\otimes_R M\cong 0.$ To prove $M$ is faithfully flat over $R,$ we would need to show that $N\cong 0.$ We have
\begin{align*}
0&\cong N\otimes_R M\\
&\cong N\otimes_R S\otimes_S M,
\end{align*}
and since $M$ is faithfully flat over $S,$ it follows that $N\otimes_R S\cong 0.$ However, unless we assume that $S$ is faithfully flat over $R,$ it may not be the case that $N\cong 0.$ We can use this to construct a counterexample to the claim that if $R\to S$ is flat and $M$ is faithfully flat over $S,$ then $M$ is faithfully flat over $R.$
Let $R = \Bbb{Z},$ $S = M = \Bbb{Q}.$ Then $M$ is clearly faithfully flat over $S,$ but we have $\Bbb{F}_p\otimes_R M\cong \Bbb{F}_p\otimes_\Bbb{Z}\Bbb{Q}\cong 0,$ although $\Bbb{F}_p$ is not $0.$ This implies that $M$ is not faithfully flat over $R.$ More generally, letting $S$ be any flat extension of $R$ which is not faithfully flat over $R$ and taking $M = S$ will provide a counterexample to this second claim.
