Overflow rule in two's complement arithmetic
Yes, if we are talking about integers. In two's complement representation with length $n$ you can only represent the integers between $-2^{n-1}$ and $2^{n-1} - 1$ (both bounds inclusive). Thus, the addition operation is not the same as the one over the integers—it is not even a proper operation because of overflows, which are left undefined; since it is a completely different operation, other rules apply (as I expect to have been previously mentioned in the book's text).
Do not mistake "overflow" for the result having a 1 carry at the left end. In example (b), for instance, the result of addition has such a carry, but it is truncated back to 4 bits, hence producing a correct result; the same happens in (d). However, an overflow occurs in both (e) and (f) because then the result is outside the representation range for 4-bit two's complement, that is, between $-2^{4-1} = -8$ and $2^{4-1} - 1 = 7$.