The discovery of this correspondence must go back to the 19th century, maybe to Dirichlet or Dedekind? And Will Jagy's answer gives a very concrete description of it.

One can also give a more conceptual description of it. Indeed, it is fairly easy, and classical, to describe the quadratic form attached to an ideal class in conceptual terms; I explain this at the end of my answer. There is also a conceptual approach to going from quadratic forms to ideal classes, but as far as I know it is more recent, and due to Melanie Wood (a version of it is mentioned in this answer).

Namely, if $Q(x,y)$ is a quadratic form over $\mathbb Z$ which is primitive, then one can consider $\mathrm{Proj} \mathbb Z[x,y]/Q(x,y)$, which turns out to be finite flat of degree $2$ over $\mathbb Z$, therefore affine, of the form $\mathrm{Spec} A$ for some quadratic order $A$. And in fact $A$ is isomorphic to $R[(D +\sqrt{D})/2]$, where $D$ is the discriminant of $Q$ (so $A$ is the quadratic order of discriminant $D$). (There is no canonical identification of $A$ with this order though; we can compose with the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ to get another isomorphism.)

Now $\mathrm{Proj}$ comes with a canonical line bundle $\mathcal O(1)$, and so this is an element of $\mathrm{Pic} A$, which can be thought of as the class group of non-zero locally principal ideals in $A$. (The usual class group when $D$ is a fundamental discriminant.)

Using the isomorphism $A \cong R[(D +\sqrt{D})/2],$ we get an element of the class group of $R[(D +\sqrt{D})/2],$ or really a pair $I,I^{-1}$ of elements, because applying the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ switches $I$ and it's inverse.

This gives a map
$$\text{ primitive quadratic forms } Q \text{ of discriminant } D \longrightarrow \text{ pairs } I, I^{-1} \text{ in the class group of } R[(D +\sqrt{D})/2].$$

To see it is a bijection, one can give an explicit inverse (as I mentioned above, this is more classical):

If $I$ is a locally principal ideal in $R[(D +\sqrt{D})/2],$ then the formula $$x \in I \mapsto N(x)/ N(I) \in \mathbb{Z}$$ defines a quadratic form $Q$ on the free rank two $\mathbb{Z}$-module $I$. (Note that $I^{-1}$ will give the same quadratic form, since formation of norms is invariant under conjugation.)

As explained below, there is a correspondence between non-zero locally principal ideals $I$ in the order of disc. $D$ and quadratic forms $Q$ of disc. $D$. The form $Q$ is naturally defined on $I$ via $x \in I \mapsto N(x)/N(I).$ So $Q(x) = n$ iff $N(x) = n N(I)$ if $N(x I^{-1}) = n,$ so representations of $n$ by $Q$ correspond to ideals in the class of $I^{-1}$ having norm $n$.

If we sum over all ideal classes, or equivalently over all $Q$, (and divide by the number of automorphisms of $Q$, which is the group of units in the order, to count ideals rather than elements that generate them) we will get the number of ideals of norm $n$.

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The discovery of this correspondence must go back to the 19th century, maybe to Dirichlet or Dedekind? And Will Jagy's answer gives a very concrete description of it.

One can also give a more conceptual description of it. Indeed, it is fairly easy, and classical, to describe the quadratic form attached to an ideal class in conceptual terms; I explain this at the end of my answer. There is also a conceptual approach to going from quadratic forms to ideal classes, but as far as I know it is more recent, and due to Melanie Wood (a version of it is mentioned in this answer).

Namely, if $Q(x,y)$ is a quadratic form over $\mathbb Z$ which is primitive, then one can consider $\mathrm{Proj} \mathbb Z[x,y]/Q(x,y)$, which turns out to be finite flat of degree $2$ over $\mathbb Z$, therefore affine, of the form $\mathrm{Spec} A$ for some quadratic order $A$. And in fact $A$ is isomorphic to $R[(D +\sqrt{D})/2]$, where $D$ is the discriminant of $Q$ (so $A$ is the quadratic order of discriminant $D$). (There is no canonical identification of $A$ with this order though; we can compose with the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ to get another isomorphism.)

Now $\mathrm{Proj}$ comes with a canonical line bundle $\mathcal O(1)$, and so this is an element of $\mathrm{Pic} A$, which can be thought of as the class group of non-zero locally principal ideals in $A$. (The usual class group when $D$ is a fundamental discriminant.)

Using the isomorphism $A \cong R[(D +\sqrt{D})/2],$ we get an element of the class group of $R[(D +\sqrt{D})/2],$ or really a pair $I,I^{-1}$ of elements, because applying the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ switches $I$ and it's inverse.

This gives a map
$$\text{ primitive quadratic forms } Q \text{ of discriminant } D \longrightarrow \text{ pairs } I, I^{-1} \text{ in the class group of } R[(D +\sqrt{D})/2].$$

To see it is a bijection, one can give an explicit inverse (as I mentioned above, this is more classical):

If $I$ is a locally principal ideal in $R[(D +\sqrt{D})/2],$ then the formula $$x \in I \mapsto N(x)/ N(I) \in \mathbb{Z}$$ defines a quadratic form $Q$ on the free rank two $\mathbb{Z}$-module $I$. (Note that $I^{-1}$ will give the same quadratic form, since formation of norms is invariant under conjugation.)