Edit: Proposition 2 is in fact false! The original answer is below in spoiler tags, but SashaP has pointed out a counterexample: if $f : k[x]\to k[x]\times k$ is the map induced by the identity on $k[x]$ and $x\mapsto 0$ from $k[x]\to k,$ and $g : k[x]\times k\to k[x]$ is the projection onto the first factor, then $gf = \operatorname{id}_{k[x]},$ but $f$ is not faithfully flat.
Notice that $B = k[x]\times k\cong k[x]\oplus k,$ and $k$ is not flat over $k[x].$ In particular, we have an exact sequence of $k[x]$-modules
$$
0\to k[x]\xrightarrow{\cdot x} k[x]\xrightarrow{\pi} k[x]/(x)\to 0
$$
and upon tensoring with $B,$ we get
$$
k[x]\oplus k\xrightarrow{\cdot x\oplus 0}k[x]\oplus k\xrightarrow{\pi\oplus\operatorname{id}_k}k[x]/(x)\oplus k\to 0,
$$
which is no longer exact on the left.
The error in the proof is that it is not true that $A\to B$ is faithfully flat if and only if for every $A$-module $N,$ the canonical map $N\to N\otimes_A B$ is injective. We must also suppose that the map $A\to B$ is flat in order to get the equivalence. Indeed, $k[x]\to k[x]\times k$ (or more generally, $A\to B\times B'$ where $B$ is faithfully flat over $A$ and $B'$ is not flat over $A$) will provide a counterexample to the original claim.
Indeed, Proposition 2 is true.
Proof: Recall that $A\to B$ is faithfully flat if and only if the canonical map $N\to N\otimes_A B$ is injective for every $A$-module $N.$ To that end, let $N$ be an $A$-module. We need to prove that $N\to N\otimes_A B$ is injective, but we know by faithful flatness of $gf$ that the composition $N\to N\otimes_A B\to N\otimes_A C$ is injective. Thus, it follows that $N\to N\otimes_A B$ must be injective. $\square$
Certainly, it need not be true that $B\to C$ be faithfully flat even if both $A\to B$ and $A\to C$ are (consider the composition $k\to k[X]\to k(X),$ for $k$ a field). So the best we can hope for is Proposition 2.
Answer: You may have seen the "flatness criteria" in Milnes "Etale cohomology" (2.7d): Let $f:A\rightarrow B$ be be a flat map of rings with $A \neq 0$. It follows $f$ is faithfully flat iff for any maximal ideal $\mathfrak{m}\subseteq A$ it follows $f(\mathfrak{m})B \subsetneq B$ is a strict ideal. In your case let
$$0 \rightarrow I:=(f_1,..,f_m) \rightarrow R[x_1,..,x_n] \rightarrow B:=R[x_i]/I \rightarrow 0$$
and assume $f:R \rightarrow B$ is flat. Use the functor $-\otimes_R R/\mathfrak{m}$ with $\mathfrak{m} \subseteq R$ a maximal ideal. It follows you get the sequence
$$I\otimes_R R/\mathfrak{m}:=(\overline{f_1},..,\overline{f_m}) \rightarrow (R/\mathfrak{m})[x_1,..,x_n] \rightarrow R/\mathfrak{m}\otimes_R B \rightarrow 0$$
hence there is an isomorphism
$$ R/\mathfrak{m}\otimes_R B \cong (R/\mathfrak{m})[x_1,..,x_n]/(\overline{f_1},..,\overline{f_m}).$$
Here the notation $(\overline{f_1},..,\overline{f_m})$ means you reduce the coefficients of the polynomials $f_i$ modulo the maximal ideal.
Question: "Is there a nice characterization of when a finitely presented R-algebra is faithfully flat in terms of the polynomials fi (or the ideal they generate)?"
The map $f$ is faithfully flat iff for any maximal ideal $\mathfrak{m} \subseteq R$ it follows the ideal
$$(\overline{f_1},..,\overline{f_m})\subseteq (R/\mathfrak{m})[x_1,..,x_n] $$
is not the unit ideal, which is a criterion on the ideal $I$.
Example: Let $A:=\mathbb{Z}[x,y]$ and let $f(x,y)\in A$ be any "non constant polynomial and let $F(x,y):=pf(x,y)-1$ where $p\in \mathbb{Z}$ is any prime number. It follows the map $f:\mathbb{Z} \rightarrow B:=A/(F)$ has the propety that
$$\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}} B \cong (0)$$
is the zero ring. If $\mathfrak{m}:=(p) \subseteq \mathbb{Z}$ it follows
the ideal $f(\mathfrak{m})B:=(p)B \subseteq B$ contains the element
$$pf(x,y) \cong -1$$
which is a unit. Hence $f(\mathfrak{m})B =B$. Hence the corresponding map
$$\pi: Spec(B) \rightarrow Spec(\mathbb{Z})$$
is not surjective: $(p) \notin Im(\pi)$.
Best Answer
If you're unaware of how one ring may be flat over another ("[b]ecause here $B[[u]]$ is a ring and so how to describe its flatness?"), it's probably a good idea to review that first:
The correct strategy to resolve your question is to recall a foundational result from commutative algebra:
This means that if we can show that $B[[u]]$ is (isomorphic to) the completion of $B\otimes_A A[[u]]$ with respect to the ideal $(1\otimes u)$, then we'll be done. This is straightforwards: the completion $R^\wedge$ is defined as the inverse limit of $R/I^n$ as $n$ varies. It is straightforwards to see that $(B\otimes_A A[[u]])/(1\otimes u)^n\cong B\otimes_A (A[u]/(u^n))\cong B[u]/(u^n)$ which clearly has inverse limit $B[[u]]$. Thus $B[[u]]$ is the completion of $B\otimes_A A[[u]]$ and is flat over it.