The point is that if $x>1$, then $x^4>|x^3|$ and $x^4>1$, so $|2x^3|\le 2x^4$ and $5\le 5x^4$, and therefore
$6x^4+|2x^3|+5\le 6x^4+2x^4+5x^4\;.$
The reason for pushing all of the terms up to multiples of $x^4$ is to allow factoring out $x^4$ in the next step: $6x^4+2x^4+5x^4=13x^4=13|x^4|$. This shows that for all $x>1$, $|6x^4-2x^3+5|\le 13|x^4|$. In other words, there is a number $x_0$ and a positive number $M$ such that
$|6x^4-2x^3+5|\le M|x^4|$
whenever $x>x_0$: specifically, $x_0=1$ and $M=13$ will work.