Shifted output of input vs output of shifted input
This is all about consequently substituting variables.
Shifting by $t_0$ means replacing the variable $t$ by $t-t_0$. If $y(t)=x(2t)$, then $y(t-t_0)=x(2(t-t_0))=x(2t-2t_0)$.

If you have a shifted input $x_2(t)=x(t-t_0)$, then the output is $y(t)=x_2(2t)$, i.e., the variable $t$ is multiplied by $2$. Replacing $t$ by $2t$ in $x(t-t_0)$ results in $x(2t-t_0)$.

So the slide is correct.

EDIT: Concerning shifting the output, imagine a triangular input $x(t)$ with its apex at $t=0$. The output is a compressed version of the input, i.e., it is also triangular with its apex at $t=0$. If the output is shifted by $t_0$ the apex is shifted to $t=t_0$. The function $x(2t-t_0)$ has its apex at $t$ satisfying $2t-t_0=0$, i.e., at $t=t_0/2$, so the correct formula for the shifted output is $x(2t-2t_0)$.