The choice of alpha < Pi/2 versus alpha <=Pi/2 is not at issue here. The range 0 <= alpha < Pi/2 corresponds to the entire sheet of the hyperboloid which has infinite extent, and therefore it won't fit in your computer screen. You will have to cut alpha off at a finite distance away from Pi/2 to produce a finite sheet. I chose to cut it off at Pi/3.
> limiting alpha to the positive domain gives one of two sheets
Extending alpha to the negative region will not produce the other sheet. Your formula is that of a single sheet of a hyperboloid. To get the other sheet replace the vector v by -v.
> Assuming v to be an elementary unit vector is too limited.
The choice v = <1,0,0> was only for illustration. You may take v to be any unit vector that you wish. If it is not a unit vector, e.g., v = <1,2,3>, then divide it by its own length, that is change it to v = <1,2,3>/sqrt(14). That normalizes it to a unit vector.