Unifying Accept States in NFA
You can unite them exactly as said in the solution, "simply
add a new final state to the original automaton, add epsilon transitions from every old final state to the new final state, and change every old final state into a regular state."

Suppose $N$ is the original NFA at a final state $F$. Suppose it can go to one of states $E_1, \cdots, E_m$ without any input if there is an epsilon transition from $F$. Suppose on input symbols $s$, $N$ changes to one of states $S_1, \cdots, S_n$.

Let $N'$ with its unique final state $U$ be the new NFA as constructed above from $N$. When $N'$ is at its regular state $F$, what will happen?


It can stay at $F$ without further input.
It can go to any one of $E_1, \cdots, E_m$ as before because of the possible epsilon transition. 
On input $s$, it will go to one of $S_1, \cdot, S_n$. 
It can go to $U$ without any further input because of the epsilon transition from $F'$ to $U$.


Note that in case 2, 3 and 4 where $N$ has moved to state $s'\in\{E_1, \cdots, E_m, S_1, \cdots, S_n\}$, $N$ can move to another state without further input if there is epsilon transition from $s'$. Furthermore, $N$ can move to still another state without further input if there is still epsilon transition. And so on. 


In particular, if $N'$ stays at $F$ on input $s$, it can go to $U$ without any further input. 

Yes, an epsilon transition functions like teleporting magically.