I'm having a hard time understanding this proof. Why is it necessary to state line 5?
If you have $A\vee B$ and wish to show it proves $C$ -- that is $A\vee B\vdash C$ -- then you may use Disjunctive Elimination.
Disjunctive Elimination is the rule that:
$$A\vee B, A\to C, B\to C \vdash C$$
So Lines 2 and 5 are where you take(assume) two halves of the disjunction $P \;\vee\; (Q\vee R)$ and aim to show that they prove the same thing. If you can do that you have:
$$\overbrace{\underbrace{P\vee (Q\vee R)}_\text{line 1.)}, \underbrace{P\to Q\vee(P\vee R)}_\text{Line 2.), 3.), 4.)}, \underbrace{(Q\vee R)\to Q\vee(P\vee R)}_\text{Line 5), 11.)} \vdash Q\vee(P\vee R)}^\text{Line 12.)}$$
Why is line 11 stated again in line 12?
Because you use Disjunctive Elimination to conclude line 11 from the assumption of $Q\vee R$.
Does this indentation help?
$\require{cancel}\begin{array}{l|l}
1.) & P∨(Q∨R) \qquad\cancelto{\text{Premise}}{\text{Assumption }}
\\
& \begin{array}{l|l}
2.) & P \qquad\text{Assumption}
\\
3.) & P∨R \qquad 2.)\text{ Disjunction Introduction}
\\\hline
4.) & Q∨(P∨R) \qquad 3.)\text{ Disjunction Introduction}
\end{array}
\\ & \begin{array}{l|l}
5.) & Q∨R \qquad\text{Assumption }
\\ & \begin{array}{l|l}
6.) & Q \qquad\text{Assumption}
\\ \hline
7.) & Q∨(P∨R) \qquad 6.)\text{ Disjunction Introduction}
\end{array}
\\ & \begin{array}{l|l}
8.) & R \qquad\text{Assumption}
\\
9.) & P∨R \qquad 8)\text{ Disjunction Introduction}
\\ \hline
10.) & Q∨(P∨R) \qquad 9)\text{ Disjunction Introduction}
\end{array}
\\ \hline
11.) & Q∨(P∨R) \qquad\text{5.), 6.), 7.), 8.), 10.) Disjunction Elimination}
\end{array}
\\ \hline
12.) & Q∨(P∨R) \qquad\text{1.), 2.), 4.), 5.), 11.) Disjunction Elimination}
\end{array}$
$$
\dfrac{
(1): P\vee (Q\vee R)
}{
\dfrac{
\dfrac{
\dfrac{
(2): P
}{
(3):P\vee R
}{\small\text{DI}}
}{
(4): Q\vee (P\vee R)
}{\small\text{DI}}
\quad,\quad
\dfrac{
\dfrac{
(5):(Q\vee R)
}{
\dfrac{
(6):Q
}{
(7):Q\vee(P\vee R)
}{\small\text{DI}}
\quad,\quad
\dfrac{
\dfrac{
(8):R
}{
(9):P\vee R
}{\small\text{DI}}
}{
(10):Q\vee(P\vee R)
}{\small\text{DI}}
}
}{
(11):Q\vee(P\vee R)
}{\small\text{DE}}
}{
(12):Q\vee (P\vee R)
}{\small\text{DE}}
}
$$
