First, for each $i = 1, 2, \dots, \frac{p-1}{2}$,
$$\frac{2i}{p} \binom{p}{2i} = \frac{(p-1)(p-2)\cdots(p-(2i-1))}{(2i-1)!} \equiv \frac{(-1)(-2)\cdots(-(2i-1))}{(2i-1)!} \equiv -1 \pmod{p}.$$
Hence
$$\begin{aligned} \sum_{i=1}^{(p-1)/2} i^{p-2} &\equiv - \sum_{i=1}^{(p-1)/2} i^{p-2} \frac{2i}{p} \binom{p}{2i} \\ &\equiv -\frac{2}{p} \sum_{i=1}^{(p-1)/2} i^{p-1} \binom{p}{2i} \\ &\equiv -\frac{2}{p} \sum_{i=1}^{(p-1)/2} \binom{p}{2i} \pmod{p} \quad \text{(by Fermat's Little Theorem.)} \end{aligned}$$
The last summation counts the even-sized nonempty subsets of a $p$-element set, of which there are $2^{p-1} - 1$.