Proposition logic, question about indirect proof. This is the solution posted by my professor.  

For the I.P from step 7) – 11), there’s a proof that can get A&A for that I.P , does this matter?
For 8), I’m assuming instead of 1, 5MP it's 5, 7MP.  
7) A        IP  
8) ~~A      7 DN  
9) ~B       8, 2 DS  
10) A       9, 1 DS  
11) A&A     7, 10 Conj  

I missed a class, so I'm not sure if this detail was went over, but I’m assuming that the indirect proof’s starting parameter can’t be used for the final P&~P conclusion right?  
Now if that’s the case, and I write something like 
6) A -> B   5 Simp  
--> 7) A        IP  
--> 8) B        6, 7 MP  
--> 9) ~~A      7 DN  
--> 10) ~B      8, 2 DS  
--> 11) B&~B    8, 10 Conj  
12) ~A      7-11 IP  

If this statement is true, aren’t I able to find A again?
12) ~A    
13) ~B      6, 12 MT  
14) A       13, 1 DS  

Wouldn’t this contradict with all the proof we just made?
And my last question is that if I ignore this, and use B for the final conclusion, 
12) ~A
13) ~B
14) B       1, 12 DS
15) B&~B    13, 14 Conj

Can I use B again? Since I already used it in the second IP
 A: Let me see what's going on.   Your professor appears to be using Classical Logic with de Morgan's Laws (DM) and modus tollens (MT) preproved, and older names for the other rules of inference than I'm used to. (Also using $\&$ for conjunction...)
He is trying to show $A\vee B, \neg A\vee\neg B\vdash \neg (A\leftrightarrow B)$ using negation introduction.
Quickly now: the premises are $A\vee B$ and $\neg A\vee \neg B$, or equivalently $\neg (\neg A\& \neg B)$ and $\neg(A\& B)$ (by deMorgan's).   Assume $A\leftrightarrow B$; that is $A\to B$ and $B\to A$.   Should we further assume $A$, then we would deduce that $B$ (via modus ponens), but this contradicts the second premise.  So we've deduced $\neg A$. Having that we can deduce $\neg B$ (via modus tollens).  However having both $\neg A$ and $\neg B$ contradicts the first premise, so we must deny the assumption.
Hence we infer $\neg (A\leftrightarrow B)$ from $A\vee B$ and $\neg A\vee\neg B$.

So...

For the I.P from step 7) – 11), there’s a proof that can get $A\&A$ for that I.P , does this matter?
For 8), I’m assuming instead of 1, 5MP it's 5, 7MP.

Yes, that is so; you found a typo.

I missed a class, so I'm not sure if this detail was went over, but I’m assuming that the indirect proof’s starting parameter can’t be used for the final P&~P conclusion right?

No, that's quite allowable, if you can do it. If assuming $P$ does allow you to infer the contradiction, $P\&\neg P$, then that is a valid indirect proof of $\neg P$.

If this statement is true, aren’t I able to find A again?
12) ~A
13) ~B$\qquad$6, 12 MT
14) A$\qquad$13, 1 DS

That works too.

Wouldn’t this contradict with all the proof we just made?

Yes it would, and that is good.
The argument is exactly that assuming $A\equiv B$ contradicts the two premises.   It doesn't matter whether you derive $A\&\neg A$ or $B\&\neg B$, or even $(\neg A\&\neg B)\&\neg(\neg A\& \neg B)$, from the assumption under the premises, as any contradiction will do.

Can I use B again? Since I already used it in the second IP

You cannot access the statement of $B$ that is quarenteened inside the subproof.   However, you may derive a new statement of $B$, if you can from accessible statements.   Such was done from the just deduced statement $\neg A$ and the first premise using disjunctive sylogism: $\neg A, A\vee B\vdash B$.
