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.
It seems to me that what Buell says about the narrow class group is not quite right (it's hard for me to say, as I don't have a copy of it). Magma tells me that in $\mathbb{Q}(\sqrt{210})$, the narrow class group is $(\mathbb{Z}/2\mathbb{Z})^{3}$ and the ideal class group is $(\mathbb{Z}/2\mathbb{Z})^{2}$. The squares in the narrow class group would be trivial.
There is a short section (less than a page) in David Cox's "Primes of the form $x^2 + ny^2$" that starts on page 128 where he discusses what happens in the real quadratic case, leaving the task of checking the details to the reader (in a sequence of exercises). Cox states that if you want to stick with the traditional notion of equivalence of forms, then the group of forms is isomorphic to the narrow class group. On the other hand, you can change the notion of equivalence, and say that $f(x,y)$ and $g(x,y)$ are equivalent if
there is a matrix $\left(\begin{matrix} p & q \\ r & s \end{matrix} \right) \in {\rm GL}(2,\mathbb{Z})$ so that $f(x,y) = \det \left(\begin{matrix} p & q \\ r & s \end{matrix}\right) g(px+qy,rx+sy)$, then the group of forms is isomorphic to the class group.
Cox also shows that the class group is a quotient of the narrow class group, and the kernel of that map has order $1$ if there is the norm of the fundamental unit in $\mathcal{O}_{K}$ is $-1$, and the kernel has order $2$ otherwise. In short the answer to your question is yes.
Best Answer
There is a concise account in the Appendix of these notes.