Denote $\alpha=\mathbb{E} u_1^2v_1^2$, $\beta=\mathbb{E} u_1v_1u_2v_2$. 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 $\alpha=\mathbb{E} w_1^2w_2^2$.

Next, we have $$1=\mathbb{E} \left(\sum_{i=1}^p w_i^2\right)^2=\alpha\cdot p(p-1)+p\cdot \mathbb{E} w_1^4,$$ thus $$\alpha=\frac{1-p\cdot \mathbb{E} w_1^4}{p(p-1)}.$$ To find $\mathbb{E} w_1^4$, we may think that $w=(w_1,\ldots,w_p)=(\xi_1,\ldots,\xi_p)/\sqrt{\sum \xi_i^2}$ where $\xi_i$ are i.i.d. standard normal. Then $$ \mathbb{E} w_1^4=\mathbb{E} \frac{\xi_1^4}{(\sum \xi_i^2)^2}=:\Theta $$ By some form of law of large numbers like Chernoff bound, the probability that $\sum \xi_i^2<p/2$ is exponentially small in $p$, that gives exponentially small contribution to the expectation $\Theta$. If $\sum \xi_i^2>p/2$, then $$\frac{\xi_1^4}{(\sum \xi_i^2)^2}<\frac{4}{p^2}\xi_1^4,$$ and since $\mathbb{E} \xi_1^4$ is a finite constant, we conclude that $\Theta=O(1/p^2)$. Thus $\alpha=1/p^2+O(1/p^3)$.

Denote $\alpha=\mathbb{E} u_1^2v_1^2$, $\beta=\mathbb{E} u_1v_1u_2v_2$. 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 $\alpha=\mathbb{E} \langle u,w\rangle^2\cdot \langle v,w\rangle^2$, as the conditional expectation clearly does not depend on $w$ (here $\langle \cdot,\cdot\rangle$ stands for the inner product). 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 $\alpha=\mathbb{E} w_1^2w_2^2$.

Next, we have $$1=\mathbb{E} \left(\sum_{i=1}^p w_i^2\right)^2=\alpha\cdot p(p-1)+p\cdot \mathbb{E} w_1^4,$$ thus $$\alpha=\frac{1-p\cdot \mathbb{E} w_1^4}{p(p-1)}.$$ To find $\mathbb{E} w_1^4$, we may think that $w=(w_1,\ldots,w_p)=(\xi_1,\ldots,\xi_p)/\sqrt{\sum \xi_i^2}$ where $\xi_i$ are i.i.d. standard normal. Then $$ \mathbb{E} w_1^4=\mathbb{E} \frac{\xi_1^4}{(\sum \xi_i^2)^2}=:\Theta $$ By some form of law of large numbers like Chernoff bound, the probability that $\sum \xi_i^2<p/2$ is exponentially small in $p$, that gives exponentially small contribution to the expectation $\Theta$. If $\sum \xi_i^2>p/2$, then $$\frac{\xi_1^4}{(\sum \xi_i^2)^2}<\frac{4}{p^2}\xi_1^4,$$ and since $\mathbb{E} \xi_1^4$ is a finite constant, we conclude that $\Theta=O(1/p^2)$. Thus $\alpha=1/p^2+O(1/p^3)$.

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 $\alpha=\mathbb{E} w_1^2w_2^2\sim (\mathbb{E} w_1^2)^2=p^{-2}$.

Denote $\alpha=\mathbb{E} u_1^2v_1^2$, $\beta=\mathbb{E} u_1v_1u_2v_2$. 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 $\alpha=\mathbb{E} w_1^2w_2^2$.

Next, we have $$1=\mathbb{E} \left(\sum_{i=1}^p w_i^2\right)^2=\alpha\cdot p(p-1)+p\cdot \mathbb{E} w_1^4,$$ thus $$\alpha=\frac{1-p\cdot \mathbb{E} w_1^4}{p(p-1)}.$$ To find $\mathbb{E} w_1^4$, we may think that $w=(w_1,\ldots,w_p)=(\xi_1,\ldots,\xi_p)/\sqrt{\sum \xi_i^2}$ where $\xi_i$ are i.i.d. standard normal. Then $$ \mathbb{E} w_1^4=\mathbb{E} \frac{\xi_1^4}{(\sum \xi_i^2)^2}=:\Theta $$ By some form of law of large numbers like Chernoff bound, the probability that $\sum \xi_i^2<p/2$ is exponentially small in $p$, that gives exponentially small contribution to the expectation $\Theta$. If $\sum \xi_i^2>p/2$, then $$\frac{\xi_1^4}{(\sum \xi_i^2)^2}<\frac{4}{p^2}\xi_1^4,$$ and since $\mathbb{E} \xi_1^4$ is a finite constant, we conclude that $\Theta=O(1/p^2)$. Thus $\alpha=1/p^2+O(1/p^3)$.