Why selection sort will require $O(N)$ time
It is simpler and more correct to say: selection sort requires $O(n)$ exchanges (and this is optimal in the worst case).

That doesn't imply $O(n)$ time, unless the other operations take negligible time. 
But from a theoretical standpoint, $O(n^2)$ is never negligible compared to $O(n)$.

In practice if a comparisons and related overhead takes $s$ seconds and an exchange and related overhead takes $e$ seconds, the total time will be

$$\frac{sn^2}2+cn=\left(\frac{sn}2+c\right)n.$$

This can be considered a linear behavior as long as $sn\ll c$. That could be true if the records are stored on a slow disk while the keys are cached in RAM.



This course plays a dangerous game with the Big-$O$ notation.