[proofplan]
Choose a linear form $L$ defining the hyperplane $H$. Since $H$ does not contain $X$, the restriction of $L$ to $X$ is not identically zero, and therefore $L^d$ is a degree-$d$ homogeneous form not vanishing identically on $X$. The zero divisor of $L^d$ on $X$ is exactly $d(H|_X)$. The ratio theorem for homogeneous forms of the same degree identifies the divisor of the rational function shown below with $(F)_0|_X-d(H|_X)$:
\begin{align*}
\frac{F}{L^d}
\end{align*}
Since this difference is principal, the two divisors are linearly equivalent.
[/proofplan]
[step:Choose a defining linear form for the hyperplane]
Since $H\subset \mathbb P_k^n$ is a hyperplane, there exists a nonzero homogeneous linear form $L\in k[x_0,\dots,x_n]_1$ such that $H=V_+(L)$. The hypothesis that $H$ does not contain $X$ means that $L$ does not restrict to the zero function on $X$. Hence $L^d\in k[x_0,\dots,x_n]_d$, and $L^d$ also does not restrict to zero on $X$. By hypothesis, $F$ is homogeneous of degree $d$ and does not restrict to zero on $X$, so $F$ and $L^d$ are homogeneous forms of the same degree, both nonzero on $X$ in the required sense.
[/step]
[step:Identify the divisor cut out by $L^d$]
By definition, the hyperplane section divisor $H|_X$ is the zero divisor on $X$ cut out by the restriction of $L$. For every prime divisor $E\subset X$, the order of vanishing satisfies
\begin{align*}
\operatorname{ord}_E(L^d)=d\,\operatorname{ord}_E(L).
\end{align*}
Therefore the zero divisor of $L^d$ on $X$ is
\begin{align*}
(L^d)_0|_X=d(L)_0|_X=d(H|_X).
\end{align*}
[guided]
The point of introducing $L^d$ is that it has the same degree as $F$, so the quotient $F/L^d$ is a well-defined rational function on the projective variety $X$. First we identify the divisor cut out by this denominator.
The hyperplane $H$ is defined by $L$, so the hyperplane section divisor $H|_X$ is, by definition, the zero divisor of the restricted homogeneous form $L$ on $X$:
\begin{align*}
H|_X=(L)_0|_X.
\end{align*}
Because $H$ does not contain $X$, the restriction of $L$ to $X$ is not identically zero, so this divisor is well-defined.
Now pass from $L$ to $L^d$. Let $E\subset X$ be any prime divisor. The coefficient of $E$ in the zero divisor of a rational function or section is its order of vanishing along $E$. Taking the $d$-th power multiplies orders of vanishing by $d$, so
\begin{align*}
\operatorname{ord}_E(L^d)=d\,\operatorname{ord}_E(L).
\end{align*}
Since this holds for every prime divisor $E\subset X$, the entire divisor satisfies
\begin{align*}
(L^d)_0|_X=d(L)_0|_X.
\end{align*}
Substituting $(L)_0|_X=H|_X$ gives
\begin{align*}
(L^d)_0|_X=d(H|_X).
\end{align*}
[/guided]
[/step]
[step:Apply the ratio theorem and conclude linear equivalence]
The hypotheses of [citetheorem:9450] apply to the homogeneous forms $F$ and $L^d$: the base field $k$ is algebraically closed; the variety $X$ is positive-dimensional, nonsingular, irreducible, and projective over $k$; the forms $F$ and $L^d$ have the same degree $d$; and neither restricts to zero on $X$. Therefore
\begin{align*}
\operatorname{div}(F/L^d)=(F)_0|_X-(L^d)_0|_X.
\end{align*}
Using the divisor identity from the previous step, this becomes
\begin{align*}
\operatorname{div}(F/L^d)=(F)_0|_X-d(H|_X).
\end{align*}
Thus $(F)_0|_X-d(H|_X)$ is a principal divisor. By the definition of linear equivalence of divisors, this proves
\begin{align*}
(F)_0|_X\sim d(H|_X).
\end{align*}
[/step]
