Why Two's Complement works
I'll stick to 8-bit quantities, but the same applies in general.
The key to understanding two's complement is to note that we have a set of finitely many (in particular, $2^8$) values in which there is a sensible notion of addition by $1$ that allows us to cycle through all of the numbers. In particular, we have a system of modular arithmetic, in this case modulo $2^8 = 256$.
Intuitively, arithmetic modulo $n$ is a system of addition (and subtraction) in which overflow and underflow cause you to "cycle back" to a value from $0$ to $n-1$. A classic example is the usual "clock arithmetic", which is to say arithmetic modulo $12$.
For example, if it is $11\!:\!00$ now, then three hours later it will be $2\!:\!00$, since $$ 11 + 3 = 14 \equiv 2 \pmod {12} $$ and similarly, if it is $1\!:\!00$, then $4$ hours ago it was $9$ since $$ 1 - 4 = -3 \equiv 9 \pmod{12} $$ Notice that subtracting $4$ hours on the clock is the same as adding $12 - 4 = 8$ hours. In particular, we could have computed the above as follows: $$ 1 - 4 \equiv 1 + 8 = 9 \pmod{12} $$ That is: when performing arithmetic modulo $n$, we can subtract $x$ by adding $n-x$.
Now, let's apply this idea modulo $256$. How do you subtract $3$? Well, by the above logic, this is the same as adding $256 - 3 = 253$. In terms of binary, we say that subtracting $00000011$ is the same as adding $$ 1\overbrace{00000000}^8 - 00000011 = 1 + \overbrace{11111111}^8 - 00000011 = 1111101 $$ and there's your two's complement.
Notably, we don't think of $1111101$ as being $253$ in our $8$-bit system, we instead consider it to represent the number $-3$. Rather than having our numbers go from $0$ to $255$ around a clock, we have them go from $-128$ to $127$, where $-x$ occupies the same spot that $n - x$ would occupy for values of $x$ from $1$ to $128$.
Note: an interesting infinite analog to the two's complement system of subtraction is that of infinite series 2-adic numbers. In particular, we can say something strange like $$ \dots 11111 = -1 $$
Let's look at a decimal example. You want to do $735-78$.
Borrow 1000 from the Number Bank; the loan is subject to no interest, but you must give back what you got as soon as you have used it.
Now consider that $$ 735-78=735+(1000-78)-1000 $$ The subtraction $1000-78$ is very easy to do: just do $9$-complement on the rightmost three digits (the missing one at the far left is, of course, $0$), getting $921+1$, so our operation now reads $$ 735-78=735+921+1-1000 $$ Since \begin{array}{rr} 735 & + \\ 921 & = \\ \hline 1656 \end{array} we can give back 1000 to the bank and add 1: $$ 735-78=656+1=657 $$
In base two it's exactly the same, with the only difference that $1$-complement (instead of $9$-complement) is very easy, because it consists in flipping the digits. You don't need the loan either, because you work on a fixed number of bits, and numbers that overflow are simply reduced forgetting the leftmost digit. So if you have to do
00101001 - 00001110
you can flip the digits in the second number and add, forgetting the leftmost bit that may become 1:
00101001 +
11110001 =
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00011010 +
1 =
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00011011