As a concrete example of having more factorizations than the class number, consider first the numbers $6, 14, 21$ in $\mathbb Z[\sqrt{-5}]$. We have
$6=2×3=(1+\sqrt{-5})(1-\sqrt{-5})$
$14=2×7=(3+\sqrt{-5})(3-\sqrt{-5})$
$21=3×7=(1+2\sqrt{-5})(1-2\sqrt{-5})=(4+\sqrt{-5})(4-\sqrt{-5})$
Note that $21$ already has three factorizations, each involving exactly two until factors. Now form their LCM and using the above factorizations render
$42=2×3×7=2×\color{blue}{21}=3×\color{blue}{14}=7×\color{blue}{6}$
where the blue numbers each have the factorizations given above. Thereby $42$ has at least five distinct factorizations (modulo units and order) in this class-2 domain.
Doubling up
At least for imaginary quadratic fields, the class number is more closely connected with how to modify the domain so as to make it UF. In the case of class 2, but not higher classes, augmenting the original integer domain with a single additional lattice in the complex plane suffices to gain UF -- if we give up closure with respect to addition.
Let's see how this works in the case of $\mathbb Z[\sqrt{-5}]$. We form the union of this set with algebraic integers having the form $a\sqrt2+b\sqrt{-10}$ with $a,b$ rational, that is the products of $\sqrt2$ with elements of $\mathbb Z[(1+\sqrt{-5})/2]$. For instance, $(\sqrt2+\sqrt{-10})/2$ whose minimal polynomial equation is $x^4+4x^2+9=0$ belongs in this second set.
When we overlay the two lattices together we lose closure with respect to addition, but the union is closed with respect to multiplication and now it is UF. Watch what happens now with the numbers I noted above:
$6=2×3=(\sqrt2)^2\cdot\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$
$6=(1+\sqrt{-5})(1-\sqrt{-5})=(\sqrt2)\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\cdot(\sqrt2)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$
$14=2×7=(\sqrt2)^2\cdot\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$
$14=(3+\sqrt{-5})(3-\sqrt{-5})=(\sqrt2)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\cdot(\sqrt2)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$
$21=3×7=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\cdot\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$
$21=(1+2\sqrt{-5})(1-2\sqrt{-5})= -1\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)\cdot(-1)\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)$
$21=(4+\sqrt{-5})(4-\sqrt{-5})=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)\cdot\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)$
The numbers $6,14,21$ are now factored into irreducibles that differ only by units and order, and so will their LCM $42$.
Class 2 imaginary quadratic fields can be overlaid with one additional lattice to form a quasiperiodic set that becomes UF with respect to multiplication. This property is connected with the fact that there are exactly two reduced quadratic forms associated with such domains (such as $x^2+5y^2$ and $2x^2+2xy+3y^2$), where the second quadratic form corresponds to the lattice we added. Classes greater than two require more extensive overlays to get similar results.
Rendering the Carlitz interpretation
KCd mentions the Carlitz interpretation of class 2: with class 2 all nonunique factorizations of a given number have the same number of non-unit irreducibles, such as $6=2×3$ and $6=(1+\sqrt{-5})(1-\sqrt{-5})$ both having two of them. This emerges from the unique factorization in the double lattice. To see how, consider the unique factorization
$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$
which has four non-units from the augmented lattice I used. To retrieve a factorization in $\mathbb Z[\sqrt{-5}]$, we must multiply together pairs of factors from the second lattice, the products of such pairs being in the original class 2 domain if the second lattice is chosen properly, for instance in this case
$(\sqrt{2})\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)=1+\sqrt{-5}$
Thus four factors from the second lattice will generate half that many, or two, when we go back into
$\mathbb Z[\sqrt{-5}]$. Then the two factorizations
$6=2×3=(1+\sqrt{-5})(1-\sqrt{-5})$
correspond to different pairings of the four unique factors from the double lattice, both of which must produce two $\mathbb Z[\sqrt{-5}]$ non-units. Similarly the unique double-lattice factorization
$6699=3×7×11×29=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$
(where natural primes $\equiv11\bmod 20$ are irreducible in the double lattice) will give five non-units when rendered back into $\mathbb Z[\sqrt{-5}]$: two from pairing the factors of $3×7=21$ which are in the second lattice, and the other three factors ($11$ and the factors of $29$) which remain unchanged:
$6699=(3×7×11)(3+2\sqrt{-5})(3-2\sqrt{-5})$
$6699=(1+2\sqrt{-5})(1-2\sqrt{-5})(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$
$6699=(4+\sqrt{-5})(4-\sqrt{-5})(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$