This sort of thing is what makes some of the structural rules of sequent calculus admissible, meaning that they can be proven (in a particular sense), rather than needing to assert them separately.
Here, leaving $A \supset B$ in the premises means that we can use it again farther up in the proof. This means that we don't need to use a contraction rule to duplicate $A \supset B$ if we want to use it twice.
If something similar is done with all the other rules, then contraction becomes admissible: whenever we have a derivation of $\Gamma, A, A, \Delta \vdash B$, we can prove that there is a derivation of $\Gamma, A, \Delta \vdash B$ too. In particular, if we have a proof that uses $A \supset B$ twice, then there's a proof that only uses it once too.
To understand why $A \supset B$ can be removed from $\Gamma, A \supset B, B \vdash C$, but not $\Gamma, A \supset B \vdash A$, let's try proving that any derivation of $\Gamma, A \supset B, B \vdash C$ can be turned into a derivation of $\Gamma, B \vdash C$.
To do this, we'll use the cut rule, which can then be eliminated.
$$
\dfrac{
\dfrac{
\dfrac{}{\Gamma, B, A \vdash B}\text{(init)}
}{
\Gamma, B \vdash A \supset B} \text{($\supset$R)} \quad
\dfrac{...}{\Gamma, A \supset B, B \vdash C}
}
{
\Gamma, B \vdash C
} \text{(cut)}
$$
The "$...$" at the top stands for the given proof of $\Gamma, A \supset B, B \vdash C$.
However, proofs of $\Gamma, A \supset B \vdash A$ can use $A \supset B$ in an essential way. The text you link gives the example of a proof of $\vdash \neg \neg (A \vee \neg A)$. Let's take a look. Just a reminder that these proof trees are constructed from the bottom up, so that may help you understand where the derivation comes from.
$\neg A$ is a shorthand for $A \supset \bot$, so we'll start by expanding that.
$$
\dfrac{
\dfrac{
\dfrac{...}
{(A \vee \neg A) \supset \bot \vdash
A \vee \neg A} \quad \quad
\dfrac{}
{(A \vee \neg A) \supset \bot, \bot \vdash \bot}
\quad \quad \text{$\bot$L}
}{
(A \vee \neg A) \supset \bot \vdash \bot
} \quad \text{$\supset$L}
}
{\vdash ((A \vee \neg A) \supset \bot) \supset \bot
} \quad \text{$\supset$R}
$$
At this point in the proof (at the "$...$"), we have a choice. We can either prove $A$ or $\neg A$ using $(A \vee \neg A) \supset \bot$. As it happens, we can use $A$ to prove $A \vee \neg A$ and then our premise to prove $\bot$, which means that we can prove $\neg A$.
Continuing the proof from there,
$$
\dfrac{
\dfrac{
\dfrac{
\dfrac
{\dfrac{}{(A \vee \neg A) \supset \bot, A \vdash A}
\quad \text{init}}
{(A \vee \neg A) \supset \bot, A \vdash A \vee \neg A}
\quad \text{$\vee$R$_1$}
\dfrac{}
{\quad (A \vee \neg A) \supset \bot, A, \bot \vdash \bot}
\quad \text{$\bot$L}
}
{(A \vee \neg A) \supset \bot, A \vdash \bot}
\quad \text{$\supset$L}
}{
(A \vee \neg A) \supset \bot \vdash A \supset \bot
}
\quad \text{$\supset$R}
}{
(A \vee \neg A) \supset \bot \vdash A \vee \neg A
}
\quad \text{$\vee$R$_2$}
$$
So we used the premise $(A \vee \neg A) \supset \bot$ twice. The second time, we had an $A$ in the context which meant that we could make a different choice when proving $A \vee \neg A$.