In order to find this parametric equation, I would do:
$$z=\sqrt{1-x²}\\x=\sqrt{1-4y²}\implies \\z = \sqrt{1-(1-4y²)} = \sqrt{1+4y²}$$
Then, if I choose $x=t$ as a parameter, I get:
$$x=t, y=\frac{\sqrt{1-t²}}{2}, z = \sqrt{1+4t²}$$
but this path is hard and I don't know how to send it from $(1,0,0)$ to $(-1,0,0)$ as requested. My book simply says that a natural parametrization for this would be:
$$x = \cos(t)\\y=\frac{\sin(t)}{2}\\z=\sin(t)\\0\le t \le \pi$$
I can see in a sort way that if I do $t_2=\cos(t)$ I will get the desired paramatrization of the book, but I'm too confused in why I should do this. Maybe a map would explain, but I'm not into it. Why the book suggests this parametrization as natural?
It is a natural parametrization because it uses the fact that the curve belongs to the cylinder $$ x^2+z^2=1. $$
Automatically when you see this equation, you should write $$ x=\cos t,\quad z=\sin t , \quad 0 \le t \le 2\pi. $$
And for $y$, you use the other equation: $$ x^2+4y^2=1 \quad \Rightarrow \quad y=\pm\frac{1}{2}\sqrt{1-x^2}=\frac{1}{2}\sin t\quad \mbox{(because $y\ge0$).} $$
You then restrict $t$ to a smaller interval if needed. Here, you want $y,z\ge 0$, so you have to take $t\in [0,\pi]$.