Trying to understand $2$'s complement Using $n$-bit integers we can usually represent anything from $0$ to $2^n-1$. If we're interpreting a number with $2$'s complement, then the first bit is the sign bit, and we represent $-x$ with $2^n-x$. So for example, normally:
$n=4$:
\begin{array}{|c|c|}
\hline
\text{Decimal} & \text{Binary}\\ \hline
0 & 0000 \\ \hline
1 & 0001 \\ \hline
2 & 0010 \\ \hline
3 & 0011 \\ \hline
4 & 0100 \\ \hline
5 & 0101 \\ \hline
6 & 0110 \\ \hline
7 & 0111 \\ \hline
8 & 1000 \\ \hline
9 & 1001 \\ \hline
10 & 1010 \\ \hline
11 & 1011 \\ \hline
12 & 1100 \\ \hline
13 & 1101 \\ \hline
14 & 1110 \\ \hline
15 & 1111 \\ \hline
\end{array}
But using $2$'s complement:
\begin{array}{|c|c|}
\hline
\text{Decimal} & \text{Binary}\\ \hline
0 & 0000 \\ \hline
1 & 0001 \\ \hline
2 & 0010 \\ \hline
3 & 0011 \\ \hline
4 & 0100 \\ \hline
5 & 0101 \\ \hline
6 & 0110 \\ \hline
7 & 0111 \\ \hline
-8 & 1000 \\ \hline
-7 & 1001 \\ \hline
-6 & 1010 \\ \hline
-5 & 1011 \\ \hline
-4 & 1100 \\ \hline
-3 & 1101 \\ \hline
-2 & 1110 \\ \hline
-1 & 1111 \\ \hline
\end{array}
Now apparently if we add two numbers using $2$'s complement we still get the right answer even if we have to truncate the leftmost bit, and this is bending my brain at the moment.
If I have two negative numbers $-x$ and $-y$, then adding them gives us $2^{n+1} - (x + y)$ which after truncating gives us $-(x+y)$.
If I have a positive number $x$ and a negative number $-y$, then adding them gives us $x + (2^n - y) = 2^n - (-x + y)$ which is the representation for $-(-x+y)$ or $x-y$.
If I have a negative number $-x$ and a positive number $y$ then adding them gives us $2^n-x + y = 2^n-(x-y)$ which is the representation for $-(x-y)$ or $y-x$.
If I have two positive numbers $x$ and $y$ then adding them gives us $x+y$.
My question:
Looking at the four cases mentioned earlier, do we normally have to handle them separately, or is it really just doing arithmetic mod $2^n$ and everything will just work itself out naturally as long as everything ($x$, $y$, as well as $x+y$) is confined to $-2^{n-1}$ through $2^{n-1}-1$? For example using $4$ bits we could not be able to represent $(-8) + (-7) = -15$ since the sum falls outside the scope?

It's the latter – arithmetic mod $2^n$ and everything works itself out naturally. And yes, we can't represent $-15$ with $4$ bits because it's outside the scope.
