How many kilometers would be the span of sky you could see between horizons, assuming a plain terrain?

Your visibility radius is the square-root of twice your height times the radius of the Earth, it comes out to about $5~ \mathrm{km}$ for a $2~ \mathrm{m}$ person (square root of $2$ times $6000~ \mathrm{km}$ times $2~ \mathrm{m}$).

This handy formula shows you that if you have a crow's nest which is $20$ $\mathrm{meters}$ above the ground, you can see $3.16$ times further, the square root of $10$, about $15~ \mathrm{km}$. If you want to see $50~ \mathrm{km}$, you need to be $200$ $\mathrm{meters}$ up, so forget about it. When you are $10~ \mathrm{km}$ above the ground, you can see $240~ \mathrm{km}$ out, this is an airplane's cruising height.

The formula comes from the law of a sphere, a sphere looks like a parabola near the top,

This is true for small $x$, it's the leading Taylor expansion of square-root. You can see as far as when the slope from your eye is tangent to the parabola. Extrapolating the tangent of a parabola from position $x$, the slope is $\frac{x}{R}$, this is the derivative, and it's a distance $x$, so it's $\frac{x^2}{R}$ in height for the line, but you are starting $\frac{x^2}{2R}$ below ground level, so the extra height for the tangent is $\frac{x^2}{2R}$

You can also just use the power of a point theorem.

I know the exact answer, but the method of approximations is more general and less intimidating.

You're right that Taylor approximation is more general, but I like using high school geometry when possible. The power of a point theorem is great here because it directly implies that the length of the tangent segment is $\sqrt{2Rh}$.

This is not the exact answer--- the exact distance in line-of-sight-through-air to the horizon is $\sqrt{2Rh+h^2}$, as you can see from the fact that there is a right-triangle formed by the radius to the meeting point of tangent and circle, the line of sight, and the radius extended through your body.