For a normal variety $X$ denote by $\operatorname{Div}(X)$ the space of Weil divisors and $\operatorname{CDiv}(X)$ the Cartier divisors. An element of $\operatorname{CDiv}(X) \otimes \mathbb{Q}$ is called a $\mathbb{Q}$-Cartier divisor. The example below is intended to show that in general $\operatorname{CDiv}(X) \subset \operatorname{CDiv}(X) \otimes \mathbb{Q}$ is a proper inclusion.
Example: Let $X$ be a hypersurface in $\mathbb{A}^3_k$ defined by $xy + z^2 = 0$. That is $X=\operatorname{Spec} \ k[x,y,z]/(xy+z^2)$.One can show that $X$ is normal (that's because the only singular point of $X$ is the origin. Now we can use Serre's criterion).
Take the closed subset $D = V(y, z)$ in $X$, then it is a prime divisor isomorphic to $\mathbb{A}^1_k$. Two claims I not understand:
I) Why is $O_X(D)$ not invertible or equivalently $D$ not a Cartier divisor?
II) Why is, on the other hand, $O_X(2D)$ a free $O_X$-Module (=invertible)?
What we know is that $D$ is a prime divisor on $X$ (a irreducible codimension one subvariety). That is for open $U \subset X$
$$\mathcal{O}_X(nD)(U):=\{ f\in K(X)\;|\; (f)\geq - nD|_U \}$$
For I) it is sufficient to show that $\mathcal{O}_X(D)(X)$ isn't generated by a single element as $k[x,y,z]/(xy+z^2)$-module. I not know how can I do it.
For II) I know that $y^{-1}, z^{-2} \in \mathcal{O}_X(2D)(X)$ and $z^{-2}=x^{-1}y^{-1}$. My guess was that $y^{-1}$ is a promissing candidate to be the single generator of $\mathcal{O}_X(2D)$ as $O_X$ module, But $x^{-1} \not \in k[x,y,z]/(xy+z^2)$ , hence my confusion if $y^{-1}$ is the right choice.
For convenience I alter your notation a bit: $k[x,y,z]=k[X,Y,Z]/(XY+Z^2)$.
As for (I) it suffices to show that the ideal $(y,z)\mathcal{O}_{X,(0,0,0)}$ is not principal. (Note that an ideal can be invertible without being generated by one element globally, so your corresponding remark is wrong.)
Suppose $(y,z)\mathcal{O}_{X,(0,0,0)}$ is a principal ideal, then one of the two elements $y$ or $z$ generates the ideal. Hence either $y=\frac{a(x,y,z)}{b(x,y,z)}z$ or $z=\frac{a(x,y,z)}{b(x,y,z)}y$, where $b\in k[X,Y,Z]$ is a polynomial with a non-zero constant coefficient. In the first case one arrives at the equation
$b(X,Y,Z)Y=a(X,Y,Z)Z+f(X,Y,Z)(XY+Z^2).$
The polynomial on the right-hand-side has no monomial of the form $cY$, $c\in k^\ast$, while the polynomial on the left-hand-side does. A similar argument holds in the second case.