I'll begin with a general remark which will be illustrated by a computation in an arbitrary order of quadratic number field.

If $\overline{I}$ contracts to $I$, i.e., if $\overline{I} \cap R = I$, then the inclusion $R \rightarrow \overline{R}$ induces an injective $R$-module homomorphism $R/I \rightarrow \overline{R}/\overline{I}$. As a result, $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ and in particular we have $N_R(I) \le N_{\overline{R}}(\overline{I})$. If for instance $I$ is a prime ideal, then $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$.

The underlying question that I fail to answer is:

**Question.** Is it always true that $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$, or at least that $N_R(I) \le N_{\overline{R}}(\overline{I})$?

**Edit.** The OP answer contains a proof that $N_R(I) \le N_{\overline{R}}(\overline{I})$ holds true for every non-zero ideal of $R$.

I will not address the above question. Instead, I'll introduce a condition on $R$ under which $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ for every non-zero ideal $I$ of $R$.

**Proposition.** If a non-zero ideal $I$ of $R$ is projective over its ring of multipliers $\varrho(I) \Doteq \{ r \in \overline{R} \, \vert \, rI \subseteq I\}$, then we have
$$
N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert.
$$

**Side note.** that $\varrho(I) = \{ r \in K \, \vert \, rI \subseteq I\}$ where $K$ denotes the field of fractions of $R$, since $R$ is Noetherian.

**Lemma 1 (OP's Claim)**. If $I$ is an invertible ideal of $R$ then
$N_{\overline{R}}(\overline{I}) = N_R(I)$.

*Proof.* First, prove the statement for a non-zero principal ideal $I$. Then decompose the $R$-module of finite length $\overline{R}/\overline{I}$ as a direct sum of its localizations with respect to the maximal ideals of $R$ [4, Theorem 2.13]. Do the same for $R/I$ and compare the cardinalities of the summands.

*Proof of the Proposition.* By Lemma 1, we have
$N_{\overline{R}}(\overline{I}) = N_{\varrho(I)}(I)$.
Hence $N_{\overline{R}}(\overline{I}) = [\varrho(I) : R][R: I] = \vert \varrho(I)/R\vert N_R(I)$.

Note that if $R$ is an order whose ideals are two-generated (e.g., an order in a quadratic field or an order whose discriminant is fourth-power free [2, Theorem 3.6]), then every non-zero ideal of $R$ satisfies the hypothesis of the above proposition, see e.g., [1], [2] and Theorem 4.1, Corollaries 4.3 and 4.4 of Keith Conrad's notes. The OP discusses similar results in his remarks and his answer.

Let $m$ be a square-free rational integer. We set $K \Doteq \mathbb{Q}(\sqrt{m})$ and denote by $\mathcal{O}(K)$ the ring of integers of the quadratic field $K$.

**Loose Claim.** Given an order $R$ of $K$ and an ideal $I \subseteq R$, we shall compute $N_{\mathcal{O}(K)}(I\mathcal{O}(K))$ as a function of $N_R(I)$ and of a binary quadratic form associated to $I$.

To do so, we introduce some notation and definitions.

Setting $$\omega = \left\{
\begin{array}{cc}
\sqrt{m} & \text{ if } m \not\equiv 1 \mod 4, \\
\frac{1 + \sqrt{m}}{2} & \text{ if } m \equiv 1 \mod 4, \\
\end{array}\right.
$$ we have $$\mathcal{O}(K) = \mathbb{Z} + \mathbb{Z} \omega$$ and any order of $K$ is of the form $\mathcal{O}_f(K) \Doteq \mathbb{Z} + \mathbb{Z} f \omega$ for some rational integer $f > 0$ [2, Lemma 6.1].
Moreover, the inclusion $\mathcal{O}_f(K) \subseteq \mathcal{O}_{f'}(K)$ holds true if and only if $f'$ divides $f$.
If $I$ is an ideal of $\mathcal{O}_f(K)$, then its ring of multipliers $\varrho(I) \Doteq \{ r \in \mathcal{O}(K) \, \vert \, rI \subseteq I\}$ is the smallest order $\mathcal{O}$ of $K$ such that $I$ is projective, equivalently invertible, as an ideal of $\mathcal{O}$ [2, Proposition 5.8].
Let us fix $f > 0$ and set $$R \Doteq \mathcal{O}_f(K), \quad \overline{R} \Doteq \mathcal{O}(K).$$

An ideal $I$ of $R$ is said to be *primitive* if it cannot be written as $I = eJ$ some rational integer $e$ and some ideal $J$ of $R$.

The main tool is the Standard Basis Lemma [5, Lemma 6.2 and its proof].

**Lemma 2.** Let $I$ be a non-zero ideal of $R$. Then there exist rational integers $a, e > 0$ and $d \ge 0$ such that $-a/2 \le d < a/2$,
$e$ divides both $a$ and $d$ and we have
$$
I = \mathbb{Z} a + \mathbb{Z}(d + e f \omega).
$$
The integers $a, d$ and $e$ are uniquely determined by $I$. We have $\mathbb{Z}a = I \cap \mathbb{Z}$ and the integer $ae$ is equal to the norm $N_R(I) = \vert R /I \vert$ of $I$. The ideal $I$ is primitive if and only if $e = 1$.

Note that, since $\mathbb{Z}a = I \cap \mathbb{Z}$, the rational integer $a$ divides $N_{K/\mathbb{Q}}(d + e f \omega)$. We call the generating pairs $(a, d + ef \omega)$ the *standard basis of $I$*. Let us associate to $I$ the binary quadratic form $q_I$ defined by $$q_I(x, y) = \frac{N_{K/\mathbb{Q}}(xa + y(d + ef\omega))}{N_R(I)}.$$

Then we have
$$eq_I(x, y) = ax^2 + bxy + cy^2$$
with $$b = Tr_{K/\mathbb{Q}}(d + ef \omega) \text { and } c = \frac{N_{K/\mathbb{Q}}(d + ef \omega)}{a}.$$
We define the *content $c(q_I)$ of $q_I$* as the greatest common divisor of its coefficients, that is $$c(q_I) \Doteq \frac{\gcd(a, b, c)}{e}.$$

**Remark.** We have $c(q_I) = \frac{\gcd(a, d, ef)}{e} = \frac{f}{f'} = \vert \varrho(I) / R \vert$ where $f'$ is the divisor of $f$ such that $\varrho(I) = \mathcal{O}_{f'}$.

**Claim.**
Let $I$ be a non-zero ideal of $R$. Then we have
$$N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert \text{ with } \vert \varrho(I)/R \vert = c(q_I).$$

*Proof.* Since $N_R(xI) = N_R(Rx) N_R(I)$ and $N_R(Rx) = N_{\overline{R}}(\overline{R}x) = \vert N_{K/\mathbb{Q}}(x) \vert$ for every $x \in R \setminus \{0\}$, we can assume, without loss of generality, that $I$ is primitive, i.e., $e = 1$. It follows immediately from the definitions that $$\overline{I} = \overline{R} I = \mathbb{Z}a + \mathbb{Z}a \omega + \mathbb{Z}(d + f \omega) + \mathbb{Z}v$$ where

$$v = \left\{
\begin{array}{cc}
f \omega^2 + d \omega & \text{ if } m \not\equiv 1 \mod 4, \\
f \frac{m - 1}{4} + (d + f) \omega & \text{ if } m \equiv 1 \mod 4. \\
\end{array}\right.$$
Now it suffices to compute the Smith Normal Form
$\begin{pmatrix}
d_1 & 0 \\
0 & d_2 \\
0 & 0 \\
0 & 0
\end{pmatrix}$
of the matrix
$A \Doteq \begin{pmatrix}
a & 0 \\
0 & a \\
d & f \\
v_1 & v_2
\end{pmatrix}$ where $(v_1, v_2)$ is the matrix of $v$ with respect to the $\mathbb{Z}$-basis $(1, \omega)$ of $\overline{R}$.
The coefficient $d_1$ is the greatest common divisor of the coefficients of $A$ and is easily seen to be $\gcd(a, d, f) = \gcd(a, b, c)$. The coefficient $d_2$ is the greatest common divisor of the $2 \times 2$ minors of $A$ divided by $d_1$ and is easily seen to be $\frac{a \gcd(c(q_I), q_I(0, 1))}{d_1} = \frac{a c(q_I)}{d_1}$. Thus $N_{\overline{R}}(\overline{I}) = d_1 d_2$ has the desired form.

[1] J. Sally and W. Vasconcelos, "Stable rings", 1974.

[2] C. Greither, "On the two generator problem for the ideals of one-dimensional ring", 1982.

[3] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with $2$-generated ideals", 1985.

[4] D. Eisenbud, "Commutative algreba with a view toward algebraic geometry", 1995.

[5] T. Ibukiyama and M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic Fields", 2014.