Starting, generally, with a commutative ring $R$ and a rank $2$ projective
module $P$ given with a quadratic form $\varphi$ we can form its Clifford algebra
$C(P)$. Its even part $S:=C^+(P)$ is then a commutative $R$ algebra of rank $2$.
If (which we may assume locally) $P=Re_1+Re_2$ we have that $S$ has basis
$1,e_1e_2=:h$ and
$h^2=e_1e_2e_1e_2=e_1(\langle e_1,e_2\rangle-e_1e_2)e_2=\langle e_1,e_2\rangle h-e_1^2e_2^2=\langle e_1,e_2\rangle h-\varphi(e_1)\varphi(e_2)$,
where $\langle-,-\rangle$ is the bilinear form associated to $\varphi$, giving an explicit
quadratic algebra over $R$. Furthermore, $C^-(P)=P$ and it thus becomes an
$C^+(P)$-module, explicitly
$h\cdot e_1=e_1e_2e_1=\langle e_1,e_2\rangle e_1-e_1^2e_2=\langle e_1,e_2\rangle e_1-\varphi(e_1)e_2$. Furthermore,
putting $L:=S/R$ we get an isomorphism $\gamma\colon \Lambda^2_RP \to L$ given by $u\land v \mapsto
\overline{uv}$ (note that $u^2=\varphi(u)$ which maps to zero in $L$). Note for future
use that, putting $[u,v]:=\gamma(u\land v)$, we also have $[[u,v]u,u]=\varphi(u)[u,v]$, where
the left hand side is well-defined as
$[([u,v]+r)u,u]=[[u,v]u,u]+r[u,u]=[[u,v]u,u]$. We have therefore constructed
from $(P,\varphi)$ a triple $(S,P,\gamma)$, where $S$ is a quadratic (i.e., projective of
rank $2$) $R$-algebra, $P$ an $S$-module projective of rank $2$ as $R$-module
and an isomorphism $\gamma\colon \Lambda^2_RP \to S/R$ of $R$-modules. Conversely, given such a
triple $(S,P,\gamma)$ there is a unique quadratic form $\varphi$ on $P$ such that
$[[u,v]u,u]=\varphi(u)[u,v]$ (where again the left hand side is well-defined). It is
easily verified that these constructions are inverse to each other.
In case $R=\mathbb Z$ I hope this gives the usual association between forms and modules over orders in quadratic number fields (but I admit shamelessly that I haven't checked it). When $R=k[t]$ I again hope this gives what we want.
In terms of algebraic group schemes and torsors we have the following situation.
If the quadratic form is perfect, we have that $S$ is an \'etale covering and
hence corresponds to an element of $H^1(\mathrm{Spec}R,\mathbb Z/2)$. It is the
image under $O_2 \to \mathbb Z/2$ of the torsor in $H^1(\mathrm{Spec}R,O_2)$
corresponding to $\varphi$. As $\mathbb Z/2$ also acts as an automorphism group of
$SO_2$ (by conjugation in $O_2$) we can use the torsor in
$H^1(\mathrm{Spec}K,\mathbb Z/2)$ to twist $SO_2$ to get $SO(\varphi)$, the connected
component of $O(\varphi)$. Torsors over this group corresponds to isomorphism classes
of pairs consisting of a rank $2$ quadratic $R$-forms $\phi$ and an $R$-isomorphism
$C^+(\phi)\simeq S$. We have an alternative description of $SO(\varphi)$: We can consider the
units $T:=S^\ast$ of $S$ as an algebraic group and have a (surjective)
norm map $T \to \mathbb G_m$ and then $SO(\varphi)$ is the kernel of this norm map.
Using the exact sequence $1\to SO(\varphi)\to T\to\mathbb G_m\to1$ we see that an $SO(\varphi)$-torsor
is given by a projective $S$-module $P$ of rank $1$ together with the
choice of an isomorphism $\Lambda^2_RP\simeq R$.
[Added] Overcoming some of my laziness I did the calculation in the integral
case ($R=\mathbb Z$ although the only thing we use is that $2$ is invertible):
Start with the quadratic form $Cx^2+Bxy+Ay^2$ (the switch between $A$ and $C$ is
to make my definition come out the same as the formula of the question). We then
have $h^2=Bh-AC$ so that $h=\frac{B-\sqrt{\Delta}}{2}$. We have $h\cdot e_1=Be_1-Ce_2$ and
$h\cdot e_2=Ae_1$. This is just an abstract module over $S$ but we can make it a
fractional ideal by mapping $e_2$ to $A$, then $e_1$ maps to $h$ so that the
fractional ideal is indeed $(A,\frac{B-\sqrt{\Delta}}{2})$.
[Added later] One could say that the association of the ideal $(A,h)$ to the
quadratic form $Cx^2+Bxy+Ay^2$ over the ring $R[h]/(h^2-Bh+AC)$ is an answer to
the question which workds in all classical cases ($R=\mathbb Z$ and
$R=k[T]$). The reason that this looks simpler than the traditional ($R=\mathbb
Z$) answer is that we let the presentation of the quadratic order depend on the
quadratic form itself. Usually we have fixed the quadratic order and want to
consider all forms with this fixed quadratic order as associated order. This
means that we should fix some normal form for the order and then express the
ideal in terms of this normal. When $R=\mathbb Z$ orders are in bijection with
their discriminants $\Delta$ which are actual integers (as the discriminant is
well-determined modulo squares of units). A normal form for the order is
$\mathbb Z+\mathbb Z\sqrt{\Delta}$ or $\mathbb Z+\mathbb Z(\sqrt{\Delta}+1)/2$ depending
on whether $\Delta$ is even or odd. A curious feature of the form
$(A,\frac{B-\sqrt{\Delta}}{2})$ of the ideal is that the distinction between the odd
and even case is not apparent. However, the crucial thing is that it expresses a
$\mathbb Z$-basis for the ideal in terms of the canonical form of the quadratic
order.
The case of $R=k[T]$ for $k$ a field of odd characteristic is somewhat deceptive
as the discriminant $\Delta=B^2-4AC$ is not quite an invariant of the quadratic order
as there are units different from $1$ that are squares (except when $k=\mathbb
Z/3$!). Hence the formula $(A,\frac{B-\sqrt{\Delta}}{2})$ is not quite of the same
nature as for the $\mathbb Z$ case as there is a very slight dependence of $\Delta$
on the form (and not just on the order). We could fix that by choosing coset
representatives for the squares as a subgroup of $k^\ast$ and then the formula for
the ideal would would take the form $(A,\frac{B-\lambda\sqrt{\Delta}}{2})$ where $\Delta$ now
has been normalised so as to have its top degree coefficient to be a coset
representative.
The case when $k$ has characteristic $2$ is more complicated. We get and order
of the form $k[T][h]/(h^2+gh+f)$ but the question of a normal form is trickier.
The discriminant of the order (in the sense of the discriminant of the trace
form) is equal to $g^2$ and as all elements of $k$ are squares we can normalise
things so that $g$ is monic. However, the order is not determined by its
discriminant. This can be seen already in the unramified case when $g=1$ when a
normal form for $f$ is that it contain no monomials of even degree and the
constant term is one of a chosen set of coset representives for the subgroup
$\{\lambda^2+\lambda\}$ of $k$. We can decide to rather arbitrarily fix a generator $H$ for
every order with $H^2=GH+F$ where one sensible first normalisation is that $G$
be monic (for which we have to assume that $k$ is perfect). Then we have that $h=H+a$, with $a\in k[T]$, and the ideal would be
$(A,H+a)$.
There is a particular (arguably canonical) choice of $H$: We assume $G$ is monic
and then can write uniquely $G=G_1G_2\cdots G_n$ with $G_i$ monic and
$G_{i+1}|G_i$. The normal form is then that $F$ have the form
$F=G_1F_1+G_1^2G_2F_2+\cdots+G_1^2G_2^2\cdots G_nF_n+G^2F'$ where $\deg F_i<\deg G_i$ and
$F'$ contain no square monomials and its constant term belongs to a chosen set of
coset representatives for $\{\lambda^2+\lambda\}$ in $k$.
One further comment relating to the classical formulas. When passing from a
fractional ideal to a quadratic form one classically divides by the norm of the
ideal (as is done in KConrad's reply). This means that the constructed form is
primitive, i.e., the ideal generated by its values is the unit ideal. Hence if
one starts with a quadratic form, passes to the ideal and then back to a
quadratic form one does not end up with the same form if the form is not
primitive. Rather the end form is the "primitivisation" where one has divided
the form by a generator for its ideal of values. This of course only makes sense
if the base ring is a PID. Even for a general Dedekind ring if one wants to work
with primitive forms one has to accept quadratic forms that take values in
general rank $1$ modules (i.e., fractional ideals).
The approach above makes another choice. It deals only with $R$-valued forms but
accepts non-primitive ones. This would seem to lead to a contradiction as
the classical construction leads to a bijection between modules and primitive
forms and the above leads to a bijection between modules and arbitrary
forms. There is no contradiction however (phew!) as the above construction leads
to smaller orders than the classical one in the non-primitive case.
Classically what one really works with (when $R$ is a PID) are lattices in $L$
(the fraction field of $S$), where a lattice $M$ is a finitely generated
$R$-submodule of $K$ containing a $K$-basis ($K$ the fraction field of $R$) for
$L$. The order one associates to $M$ is the subring of $L$ stabilising $M$. When
the quadratic form $\varphi$ is non-primitive $C^+(P)$ is not equal to this
canonically associated order but is a proper suborder by it. Dividing by a
generator for the ideal of values gives us a primitive form for which $C^+(P)$
is equal to the canonical order.
Finally there is a particular miracle that occurs for lattices in quadratic
extensions (of the fraction field of a Dedekind ring $R$), $M$ as a module over its stabilisation order is
projective. This is why classes of primitive forms with fixed order are in
bijection with the class group of the order.
I just wanted to remark that if $p$ is a prime such that
$\ell$ splits in $F = \mathbb{Q}(\sqrt{-p})$ for all $\ell \le N$,
then one may prove the existence of $N$ consecutive integers which
are norms of integers in $\mathcal{O}_F$, providing one is willing
to assume a standard hypothesis about prime numbers, namely,
Schinzel's Hypothesis H.
First, note the following:
Lemma 1: If $C$ is an abelian group of odd order, then there exists a finite
(ordered) set $S = \{c_i\}$ of elements of $C$ such that every element in $C$ can be
written in the form
$\displaystyle{\sum \epsilon_i \cdot c_i}$
where $\epsilon_i = \pm 1$.
Proof: If $C = A \oplus B$, take $S_C = S_A \cup S_B$. If $C
= \mathbb{Z}/n \mathbb{Z}$ then take $S = \{1,1,1,\ldots,1\}$ with
$|S| = 2n$.
Let $C$ be the class group of $F$. It has odd order, because
$2$ splits in $F$ and thus $\Delta_F = -p$.
Let $S$ be a set as in the lemma. Let $A$ denote an ordered
set of distinct primes $\{p_i\}$ which split in $\mathcal{O}_F$ such
that one can write $p_i = \mathfrak{p}_i \mathfrak{p}'_i$ with
$[\mathfrak{p}_i] = c_i \in C$, where $c_i$ denotes a set of elements
whose existence was shown in Lemma 1.
Lemma 2: If $n$ is the norm of some ideal $\mathfrak{n} \in \mathcal{O}_F$,
and $n$ is not divisible by any prime $p_i$ in $A$, then
$$n \cdot \prod_{A} p_i$$
is the norm of an algebraic integer in $\mathcal{O}_F$.
Proof: We may choose
$\epsilon_i = \pm 1$ such that
$\displaystyle{[\mathfrak{n}] + \sum \epsilon_i \cdot c_i = 0 \in C}$.
By assumption, $[\mathfrak{p}_i] = c_i \in C$ and thus
$[\mathfrak{p}'_i] = -c_i \in C$. Hence the ideal
$$\mathfrak{n}
\prod_{\epsilon_i = 1}
{ \mathfrak{p}}
\prod_{\epsilon_i = -1}
\mathfrak{p}'$$
is principal, and has the desired norm.
By the Chebotarev density theorem (applied to the Hilbert class field
of $F$), there exists a set $A$ of primes as above which avoids any fixed finite set of
primes. In particular, we may
find $N$ such sets which are pairwise distinct and which contain
no primes $\le N$. Denote these sets by $A_1, \ldots, A_N$.
By the Chinese remainder theorem, the set of integers $m$ such that
$$m \equiv 0 \mod p \cdot (N!)^2$$
$$m + j \equiv 0 \mod \prod_{p_i \in A_j} p_i, \qquad 1 \le j \le N$$
is of the form $m = d M + k$ where $0 \le k < M$, $d$ is arbitrary, and $M$
is the product of the moduli.
Lemma 3: Assuming Schinzel's Hypothesis H, there exists infinitely many integers
$d$ such that
$$ P_{dj}:= \frac{dM + k + j}{j \cdot \prod_{p_i \in A_j} p_i}$$
are simultaneouly prime for all $j = 1,\ldots,N$.
Proof: By construction, all these numbers are coprime to $M$ (easy check).
Hence, as $d$ varies, the greatest common divisor of the product of these
numbers is $1$, so Schinzel's Hypothesis H applies.
Let $\chi$ denote the quadratic character of $F$. Note that
$dM + k + j = j \mod p$, and so
$\chi(dM + k + j) = \chi(j) = 1$ (as all primes less than $N$ split in $F$).
Moreover, $\chi(p_i) = 1$ for all
primes $p_i$ in $A_j$
by construction. Hence
$\chi(P_{dj}) = 1$. In particular, if $P_{dj}$ is prime, then
$P_{dj}$ and $j \cdot P_{dj}$ are norms of (not necessarily principal) ideals in
the ring of integers of $F$. By Lemma 2, this implies that
$$dM + k + j = j \cdot P_{dj} \prod_{p_i \in A_j} p_i$$
is the norm of some element of $\mathcal{O}_F$ for all $j = 1,\ldots, N$.
One reason to think that current sieving technology will not be sufficient
to answer this problem is the
following: when Sieving produces a non-trivial lower bound, it usually
produces a pretty good lower bound. However, there are no good (lower) bounds known for the following problem: count the number of integers $n$
such that $n$, $n+1$, and $n+2$ are all sums of two squares. Even for the
problem of estimating the number of $n$ such that $n$ and $n+1$ are both sums of squares is tricky - Hooley implies that the natural sieve does not give lower bounds
(for reasons analogous to the parity problem). Instead,
he relates the problem to sums of the form
$\displaystyle{\sum_{n < x} a_n a_{n+1}}$ where
$\sum a_n q^n = \theta^2$ is a modular form. In particular, he
implicitly uses
automorphic methods which won't work
with three or more terms.
Best Answer
Such a $C$ does not exist, if I got it right.
My example is the function $f(x,y)=(2x+1)(5y+1)$. An integer is represented by $f$ iff it can be written as the product of an odd integer and a number which is 1 mod 5. So for example it is not hard to check that 2 cannot be written in this way (the odd number had better be $\pm1$, and neither of $\pm2$ are 1 mod 5). However a tedious construction using the Chinese Remainder Theorem gives arbitrarily large sequences of consecutive integers which are representable by $f$.
Let me explain a bit more about the CRT argument. Clearly any positive integer $N$ can be written as the product $M2^r$ of an odd integer and a power of two. If furthermore this odd integer $M$ is divisible by three distinct primes, one of which is 2 mod 5, one of which is 3 mod 5 and one of which is 4 mod 5, then moving one of these factors $p$ from $M$ to the power of two gives $N=(M/p)(p2^r)$, and if $2^r$ isn't 1 mod 5 then there will be some $p$ such that $p2^r$ will be 1 mod 5, so we have our decomposition of $N$ which is hence representable by $f$.
So it suffices to find arbitrarily large strings of integers each of which has a prime factor which is 2 mod 5, a prime factor which is 3 mod 5 and a prime factor which is 4 mod 5 (in fact by using sign changes one can even get away with primes which are 2 mod 5). But now this is easy: enumerate the primes which are 2 mod 5, 3 mod 5 and 4 mod 5, let C_n be the product of the n'th 2-mod-5, the n'th 3-mod-5 and the n'th 4-mod-5, and now choose a big positive $N$ with $N=0$ mod C_1, $N+1=0$ mod $C_2$,... $N+10^6=0$ mod $C_{10^6+1}$ (such an $N$ exists by CRT) and there you have it.