Orbital speed is (vector) sum of tangential and normal speed? Orbital speed definition in wiki does not state clearly - it is just tangential speed component or square root of squares of normal and tangential speed (full speed vector).

When we say that moon orbital speed is 1 km per sec, we don't know its tangential speed (speed along its trajectory), right? We need some complicated math to calculate it (because of normal speed component inside that 1 km/sec orbital speed)?
"For any object moving through space, the velocity vector is tangent to the trajectory."(Citing https://en.m.wikipedia.org/wiki/Orbital_state_vectors).

Therefore, it's both of these components, there is no normal component1 (relative to the central body)... the orbital speed represents the intensity of the orbital velocity which is always tangent to the ellipse (or parabola/hyperbola) the body in orbit describes.

The Moon's orbital speed being 1 km/s means it moves along an ellipse around the Earth at a speed of 1 km/s (this speed is roughly constant, so that ellipse can sometimes be approximated to a circle).

1However you can separate orbital motion into two components normal to each other (the most popular such separation is to the radial component and one normal to it, but this is unnecessary for circular orbits where there is no radial velocity, only radial acceleration which causes the curvature of the trajectory), and in that case the orbital speed will indeed be the square root of sum of the squares of those two. 

$$v^2 = v_{||}^2 + v_{\perp}^2$$

But in two-body Kepler orbits there is no velocity component normal to the plane of the orbit, which is the plane containing the $\mathbf{v_{||}}$ and $\mathbf{v_\perp}$, which is also the plane containing $\mathbf{r}$ and $\mathbf{v}$.