Why is bit error rate = symbol error rate / number of bits per symbol in QPSK
In general, the bit error probability depends on the symbol error probability and the coder, i.e., the mapping between bits and symbols. The formula you refer to

$$P[\textrm{bit error}]\approx\frac{1}{M}P[\textrm{symbol error}]\tag{1}$$

with $M$ being the number of bits per symbol ($M=2$ for QPSK) is an approximation for a certain type of coder under the assumption that the SNR is relatively high.

In $(1)$ it is assumed that the system implements a Gray code, where nearest neighbor symbols only differ by one bit. So if the SNR is high, most symbol errors occur due to confusing a symbol with one of its nearest neighbors, which causes only one out of $M$ bits to be wrong.