1) The Cartier divisor corresponding to the Weil divisor $1.P$ is given by the pair $s=\lbrace (U,z),(V, 1) \rbrace$ described as follows:
$\bullet$ $U$ is an open neigbourhood of $P$ and $z$ is a regular function on $U$ whose sole zero is $P$ with multiplicity one.
$\bullet \bullet$ $V=X\setminus \lbrace P\rbrace $ and of course $1$ is the constant function equal to $1$ on $V$.
(This pair determines a section $s\in \Gamma(X,\mathcal K^* _X/\mathcal O^*_X)$ if you unravel what it means to be a section of a quotient sheaf.)
2) The invertible sheaf corresponding to $P$ is the sheaf denoted by $\mathcal O(P)$.
Its $k$-vector space of sections $\Gamma(W,\mathcal O(P))$ over an open subset $W\subset X$ consists of those rational functions $f\in Rat(W)=Rat(X)$ regular on $W$ except perhaps at $P$, where $f$ is allowed a pole of order at most $1$.
Let's understand the subsequent arguments of the proof in steps. I will use the notation in your link's answer (, in which $T_x = \mathrm{Spec}\mathcal{O}_x$, and $X^{(1)}$ is the set of 1-codimensional points of $X$).
- The principal divisor $(f_x)$ on $X$ has the same restriction to $\mathrm{Spec}\mathcal{O}_x$ as $D$, hence they differ only at prime divisors which do not pass through $x$.
Recall that $D_x=\sum_{y\in X^{(1)}\cap T_x}n_y\cdot(\overline{\{y\}} \cap T_x)$, and that as a principal divisor in the $X$, $(f_x)=\sum_{y\in X^{(1)}}v_y(f_x)\cdot \overline{\{y\}}$, where $v_y$ is the valuation on the prime divisor $\overline{\{y\}}$.
Then note that the valuation of $\overline{\{y\}}$ is just the valuation of $\overline{\{y\}}\cap T_x$, since they have the same local rings on $y$. Thus the restrictions of $D_x$ and $(f_x)$ are both equal to $\sum_{y\in X^{(1)}\cap T_x}v_y(f_x)\cdot(\overline{\{y\}}\cap T_x)$, and globally they differ just at those points not in the $T_x$. But it's clear that a prime divisor doesn't pass through $x$, if and only if its generic point is not in the $T_x$.
- There are only finitely many of these which have a non-zero coefficient in $D$ or $(f_x)$, so there is an open neighbourhood $U_x$ of $x$ such that $D$ and $(f_x)$ have the same restriction to $U_x$.
That's really clear because $T_x$ is just the intersection of all open neighbourhoods of $x$, and replacing $T_x$ by a neighbourhood $U_x$ won't change the coefficients in both divisors.
- Covering $X$ with such open sets $U_x$, the functions $f_x$ give a Cartier divisor on $X$. Note that if $f$,$f'$ give the same Weil divisor on an open set $U$, then $f/f'\in\Gamma(U,\mathcal{O}^*)$, since $X$ is normal.
The second sentence needs to be verified. $(f)=(f')$ implies that $(f/f')=0$, whence in each $y\in X^{(1)}$, $f/f'$ is a unit of $\mathcal{O}_y$. For every affine open subset of $X$, say $\mathrm{Spec}A$, I claim that $f/f'$ is a global section on it, i.e. $f/f'\in A$. Indeed, each prime ideal $\mathfrak{p}$ of $A$, of height $1$, corresponds to a point of codimension $1$ in $X$, and thus $f/f'\in A_\mathfrak{p}$. Since $A$ is a normal noetherian domain, by commutative-algebra theories, we have
$$A=\bigcap_{\mathrm{ht}\mathfrak{p}=1}A_\mathfrak{p}.$$
Therefore $f/f'\in A$. Then given an affine open covering $\{U_i\}$ of $U$, $(f/f')|_{U_i}\in\Gamma(U_i,\mathcal{O})$, so one can glue them into a section in $\Gamma(U,\mathcal{O})$. Actually this section is just $f/f'$ because they must be the same in $\mathscr{K}$. Thus $f/f'\in\Gamma(U,\mathcal{O})$. Similarly, $f'/f$ is also in the $\Gamma(U,\mathcal{O})$. So we have $f/f'\in\Gamma(U,\mathcal{O}^*)$.
Best Answer
Imagine the doubly-infinite (singular) cone $x^2+y^2=z^2$ in 3-space. If we intersect with the plane $x=z$, we get a single line $x=z, y=0$, a Weil divisor. But it cannot be Cartier, since it is not locally principal at the origin; any polynomial vanishing on this line will vanish with multiplicity, or else vanish on more than one component.
Algebraically speaking, the corresponding ideal of the local ring is not principal (though its square is).