What they are basically telling is that the gravitational potential at an infinite distance from the fixed mass is zero. The total energy of the satellite would be K.E. + P.E.
Since, $$V(r)=-\int{E(r)}$$
$$V(r)=\frac{-GM}{r}$$ With M being the mass of the fixed object. As r tends to infinity, it is clear that the potential of the system rapidly decreases, so it is assumed to be zero. (extremely negligible.)
The orbital velocity of the object is $v_o$. and therefore its kinetic energy is $\frac{mv_o^2}{2}$. In this type of simple orbit scenario, it is always true that $P.E=-2(K.E)$.
Therefore the total energy is $K.E-2(K.E)=-K.E.$ which is $\frac {-mv_o^2}{2}$
In a general satellite planet system, the satellite travels with an orbital velocity. Since the orbital velocity in this case makes the object travel in a circular orbit, it must have centripetal acceleration.
Therefore, the centripetal acceleration $\frac{mv_o^2}{r}=\frac{GMm}{r^2}$, $v_o=(\frac{GM}{r})^{1/2}$. Substituting in the K.E. equation, you have $K.E=\frac{GMm}{2r}$ and the gravitational potential energy at the point, by definition is $\frac{-GMm}{r}$ and hence P.E.=-2(K.E)