It is not true that a hyperplane corresponds to the invertible sheaf $\mathcal O(1)$. What is true, is that the class of a hyperplane corresponds to the invertible sheaf $\mathcal O(1)$.
Here's (a sketch of) the correspondence. The sheaf $\mathcal O(1)$ is generated by global sections, and the sections correspond to hyperplanes in $X$. Recall that the sheaf is generated by $x_0,\cdots,x_n$. Now any two hyperplanes represent the same non-trivial class in $\mathrm{Cl}(X)$, because if $H_1$ is the zero set of $a_0x_0+\cdots+a_nx_n$, and $H_2$ is the zero set of $b_0x_0+\cdots+b_nx_n$, then $H_2-H_1=((a_0x_0+\cdots+a_nx_n)/(b_0x_0+\cdots+b_nx_n))$, so they are equal in $\mathrm{Cl}(X)$.
So the class of a hyperplane corresponds to all global sections of $\mathcal O(1)$, and hence we can identify these two sets (for our purposes). And since $\mathcal O(1)$ is generated by global sections, they determine $\mathcal (X)$.
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Here's an example of what is ment by "generated by a hyperplane section". Let $D$ be the divisor defined by the ideal sheaf $(x^3) \subseteq \mathcal O_X$. This corresponds to the divisor $(x^3)=3(x)$, i.e. 3 times a hyperplane. Similarly, every divisor on $\mathbb P^n$ is determined by its degree.
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
Your guess is correct. The geometric (or function-theoretic) meaning is that sections of $\omega(D)$ are sections of $\omega$ that are allowed to have simple poles along $D$.