When you run out of places to keep the largest digit, the value overflows back to its smallest value. e.g. 111 + 001 = 1000. However, if you only have 3 bits, then it goes to 0.
For 2’s Complement, it’s a bit more complicated. e.g. 1100 + 1100 = 11000 → 1000 1100 → 1011 → 0100 4 1000 → 0111 → 1000 8 so even though the bits do overflow in the basic binary sense, they are serving their 2’s Complement purpose as expected