Note: First of all note that if $R$ is any commutative ring. For any prime ideal $P$ of $R$, the localization $R_P$ is flat $R$-module. Set $P := \bigoplus_{P\in Spec(R)} R_P$ is also flat. We claim that $P$ is faithfully flat. Indeed, if $M$ is any
$R$-module such that $P \otimes M = 0$, then $$0 = (\bigoplus_{P\in Spec(R)} R_P)\otimes M = \bigoplus_{P\in Spec(R)} (R_P\otimes M) = \bigoplus_{P\in Spec(R)} M_P.$$ Then
each localization $M_P = 0$. It is well known that this implies that $M = 0$, so
$P$ is faithfully flat $R$-module.
(Note that some of above equalities are actualy isomorphism)
Example: Set $R:= \mathbb{Z}$ and $P := (p)$ where p is a prime, then for any
prime $l\ne p$, the nonzero $\mathbb{Z}$-module $\mathbb{Z}/l \mathbb{Z}$ localizes to zero with respect to
$P$. If $P = (0)$, any torsion $\mathbb{Z}$-module localizes to zero with respect to $P$. This shows that no $\mathbb{Z}_P$ is faithfully flat over $\mathbb{Z}$. But by above note $$P := \bigoplus_{P\in Spec(\mathbb{Z})} \mathbb{Z}_P$$ is faithfully flat over $\mathbb{Z}$.
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
Formal properties
The tensor product of two faithfully flat modules is faithfully flat.
If $M$ is a faithfully flat module over the faithfully flat $A$-algebra $B$, then $M$ is faithfully flat over $A$ too.
An arbitrary direct sum of flat modules is faithfully flat as soon as at least one summand is.
(But the converse is false: see caveat below.)
Algebras
An $A$-algebra $B$ is faithfully flat if and only if it is flat and every prime ideal of $A $ is contracted from $B$, i.e. $\operatorname{Spec}(B) \to\operatorname{Spec}(A)$ is surjective.
If $A\to B$ is a local morphism between local rings, then $B$ is flat over $A$ iff it is faithfully flat over $A$.
Caveat fidelis flatificator
a) Projective modules are flat, but needn't be faithfully flat. For example $A=\mathbb Z/6=(2)\oplus (3)$ shows that the ideal $(2)\subset A$ is projective, but is not faithfully flat because $(2)\otimes_A \mathbb Z/2=0$.
b) A ring of fractions $S^{-1}A$ is always flat over $A$ and never faithfully flat [unless you only invert invertible elements, in which case $S^{-1}A=A$].
c) The $\mathbb Z$-module $\oplus_{{{\frak p}}\in \operatorname{Spec}(\mathbb Z)} \mathbb Z_{{\frak p}}$ is faithfully flat over $\mathbb Z$. All summands are flat, however none is faithfully flat.