Why computation of $u*v^3*(u*v^7)^{(p-5)/8}$ is suggested instead of $(u/v)^{(p+3)/8}$
Note that
$$\frac{7(p-5)}8+3=\frac{7p-11}8\equiv \frac{7p-11}8-p-1\equiv -\frac{(p+3)}8\pmod{p-1}.$$
Similarly
$$\frac{p-5}8+1=\frac{p+3}8$$
And so $v^3(v^7)^{\frac{p-5}8}\equiv v^{-\frac{(p+3)}8}\pmod p$.
Likewise $u(u)^{\frac{p-5}8}\equiv u^{\frac{p+3}8}$ as required.
The reason is to save a modular division which is quite a pricey operation.