The extra minus sign which distinguishes the transformation behavior of a pseudovector from that of a vector arises in two different situations - an active transformation which sends a vector $A\rightarrow -A$, and an orientation-reversing passive transformation.
The cross-product of two vectors $A$ and $B$ is usually defined component-wise as
$$C_i = \sum_{j,k=1}^3\epsilon_{ijk} A_j B_k \tag{$\star$}$$
In other words, the components of $C$ in some basis are to be computed from the components $A$ and $B$ in that basis via $(\star)$. As a result, if we make a passive transformation to a new basis $(\hat x',\hat y',\hat z')$, the components of $A\times B$ in the new basis are given by
$$ \overline C_i = \sum_{j,k=1}^3\epsilon_{ijk} \overline A^i \overline B^j\tag{$\star\star$}$$
where the line denotes the components in the new basis. It's not hard to show that under a transformation $T$ which transforms the components of ordinary vectors as $A^i\mapsto \overline A^i = T^i_{\ \ j} A^j$, the components of $C$ transform as $C^i \mapsto \overline C^i =\mathrm{det}(T) T^i_{\ \ j} C^j$. This extra factor of $\mathrm{det}(T)$ means that $C$ picks up an extra minus sign under orientation-reversing transformations.
The true nature of pseudovectors can be understood by constructing the exterior algebra over $\mathbb R^3$, denoted $\bigwedge(\mathbb R^3)$. Formally, the idea is that we take our basis $\hat x,\hat y, \hat z$ and define the so-called bivectors $\hat x \wedge \hat y, \hat y \wedge \hat z$, and $ \hat z \wedge \hat x$, where $\wedge$ is understood to be antisymmetric so e.g. $\hat x \wedge \hat y = - \hat y \wedge \hat x$ and $\hat x \wedge \hat x = 0$. We further introduce the trivector $\hat x \wedge \hat y \wedge \hat z$.
The vector space of linear combinations of these objects is the exterior algebra $\bigwedge(\mathbb R^3)$. It includes:
$$ a\tag{scalars}$$
$$ a\hat x + b \hat y + c \hat z \tag{vectors}$$
$$ a (\hat x\wedge\hat y) + b(\hat y \wedge \hat z) + c (\hat z \wedge \hat x) \tag{bivectors}$$
$$ a(\hat x \wedge \hat y \wedge \hat z) \tag{trivectors}$$
To standardize the language and allow for easier generalization, we call these objects $k$-blades. Scalars are $0$-blades, vectors are $1$-blades, bivectors are $2$-blades, and trivectors are $3$-blades.
There are two important features to note here. First, because $\mathbb R^3$ is a 3-dimensional vector space, there is only one linearly independent $3$-blade and there are no non-zero $k$-blades for $k>3$; the antisymmetry property of $\wedge$ means that if we repeat a vector (e.g. $\hat x \wedge \hat y \wedge \hat z \wedge \hat x$) then the result must be zero.
The second thing to note is that there are the same number of scalars as $3$-blades, and the same number of $1$-blades as $2$-blades. For a general $n$-dimensional vector spaces, there are the same number of $k$-blades as $(n-k)$-blades - namely ${n\choose{k}}=\frac{n!}{k!(n-k)!}$ of them.
For $\bigwedge(\mathbb R^3)$, this suggests that we can define a pairing between $k$-blades and $(3-k)$-blades, but to do so we need one more piece of information - we need to choose a special $3$-blade $\omega$ which is constructed by wedging together our orthonormal basis vectors. The standard choice is $\omega := \hat x \wedge \hat y \wedge \hat z$. Once we have this, we define the Hodge dual $\star$ of e.g. $\hat x$ via
$$\hat x \wedge (\star \hat x) = \omega \implies \star \hat x = \hat y \wedge \hat z$$
Similarly,
$$\star 1 = \hat x \wedge \hat y \wedge \hat z$$
$$\star \hat x = \hat y\wedge\hat z \qquad \star \hat y = \hat z \wedge \hat x \qquad \star \hat z = \hat x \wedge \hat y$$
$$\star (\hat y\wedge \hat z) = \hat x \qquad \star (\hat z \wedge \hat x) = \hat y \qquad \star (\hat x \wedge \hat y) = \hat z$$
$$\star (\hat x \wedge \hat y \wedge \hat z) = 1$$
This extends to all elements of $\bigwedge(\mathbb R^3)$ by linearity.
Important note: the orientation $\omega$ should be understood as part of the definition of the basis, i.e. its order; $(\hat x,\hat y,\hat z)$ is different from $(\hat y,\hat x,\hat z)$ despite using the same three unit vectors. As a result, when we change basis $(\hat x,\hat y,\hat z)\rightarrow (\hat x',\hat y',\hat z')$, it is understood that we also change the orientation $\omega = \hat x \wedge\hat y \wedge \hat z \rightarrow \hat x' \wedge \hat y' \wedge \hat z' = \omega'$.
Having defined all of this machinery, we can understand the cross product of two vectors to be the Hodge dual of their wedge product, i.e.
$$A \times B := \star (A\wedge B)$$
For example, let $A = 3\hat x + \hat z$ and $B = \hat x + \hat y$. Their wedge product is
$$A\wedge B = (3\hat x + \hat z)\wedge(\hat x + \hat y) = 3\underbrace{\hat x \wedge \hat x}_{=0} + 3\hat x \wedge \hat y + \hat z \wedge \hat x + \hat z \wedge \hat y$$
$$= 3\hat x \wedge \hat y + \hat z \wedge \hat x \color{red}{-} \hat y \wedge \hat z$$
Taking the hodge dual using the pairings defined above yields
$$\star (A\wedge B) = 3\star(\hat x \wedge \hat y) + \star(\hat z \wedge \hat x) - \star(\hat y \wedge \hat z) = 3 \hat z + \hat y - \hat x$$
which is indeed the cross product $A\times B$ as expected.
This gives us the ability to understand the cross-product at a much deeper level. Its easy to see that under the inversion $(\hat x,\hat y,\hat z)\rightarrow (\hat x',\hat y',\hat z')=(-\hat x,-\hat y, - \hat z)$, the components of a vector (1-blade) $A$ change their sign, but the components of the 2-blade $A\wedge B$ do not; this is because, for example,
$$\hat x \wedge \hat y \rightarrow \hat x' \wedge \hat y' = (-\hat x)\wedge(-\hat y) = \hat x \wedge \hat y$$
Inheriting a new orientation $\omega'=\hat x'\wedge\hat y'\wedge \hat z'$, we find that the components of $A\times B := \star(A\wedge B)$ are invariant, and thus differ from the transformation behavior of the components of a vector by a sign flip.
