Time variance of $y(t)=x(2t)$
The signals plotted on the figure are correct. 

As your provided link already shows: $$y_d(t) = \mathcal{T}\{x_d(t)\} = \mathcal{T}\{x(t-d)\} = \mathcal{T}\{x(t-2)\} = x(2t-2) \not= y(t-2) $$ where $$ ~ y(t)=\mathcal{T}\{x(t)\} = x(2t) \longrightarrow y(t-2) = x(2(t-2)) = x(2t-4) $$ 

Looking at the progression on the provided figure, $x_1(t)$ is a pulse of domain $~ [-2,2] ~$ then $y_1(t)=\mathcal{T}\{x_1(t)\} = x_1(2t)$ will be a pulse of domain $~ [-1,1] ~$ which is obtained by compressing $x(t)$ by $2$ about the origin $~t=0~$

$x_2(t)$ is obtained from $x_1(t)$, delaying it by $2$ units to the right, hence $x_2(t)$ has a domain of $~[0,4]$.
Therefore $y_2(t)=\mathcal{T}\{x_2(t)\} = x_2(2t)$ will be a pulse of domain $~ [0,2] ~$ which is obtained by compressing $x_2(t)$ by $2$ about the origin $~t=0~$

On the other hand shifting the pulse of $y_1(t)$ with the domain $[-1,1]$ by $2$ units to the right creates a new pulse with domain $[1,3]$ which is not equivalent to $[0,2]$; the domain of $y_2(t)$, from which we deduce, also graphically, that $y_1(t-2) \not = y_2(t)$