Denote $\alpha=\mathbb{E} u_1v_1$, $\beta=\mathbb{E} u_1v_1u_2v_2$. Then by the symmetry and linearity of expectation we have $$f(m):=\mathbb{E} (u_1v_1+\ldots+u_mv_m)^2=m\alpha+m(m-1)\beta.$$ We have $f(p)=0$, thus $\beta=-\alpha/(p-1)$, and $f(m)=\alpha m(p-m)/(p-1)$.
It remains to bound $\alpha$. Choose a vector $w=(w_1,\ldots,w_p)\in \mathcal{S}^{p-1}$ independent of $u,v$, then $\alpha=\mathbb{E} \langle u,w\rangle\cdot \langle v,w\rangle$, as the conditional expectation clearly does not depend on $w$. On the other hand, the conditional expectation does not depend on the pair $u,v$ of orthogonal vectors $u,v$, thus we may take $u=(1,0,\ldots,0)$, $v=(0,1,\ldots,0)$, and $\alpha=\mathbb{E} w_1^2w_2^2$. Gaussian interpretation of the distribution on sphere shows that $w_1^2,w_w^2$ are almost independent, thus $\alpha=\mathbb{E} w_1^2w_2^2\sim (\mathbb{E} w_1^2)^2=p^{-2}$.