Please help me to understand Step (2),(3),(4) and (6) of this theorem. I know mean Value theorem of calculus and here in first step they use it. This is something that mvt implies and I don"t know how we can write this inequality. I will be grateful if anyone can elaborate me all these steps.
Thanks a lot.

For $\psi\in\mathcal{D}_Q$, you have $\psi(0)=0$ and so for $i\in\{1,\dots,n\}$ and $x=(x_1,\dots,x_i,\dots,x_n)$ the mean value theorem (in the $i$-th variable) implies $$ |\psi(x)| = |\psi(x) - \psi(0) | = |(x_i-0)(D^i\psi)(x_1,\dots,\xi_i,\dots,x_n)| \leq \max_{x\in Q} |D^i\psi|. $$ Now by the fundamental theorem of calculus, you have $$ \psi(y_1,\dots,y_i,\dots,y_n) = \int_0^{y_i} (D^i\psi)(y_1,\dots,y_{i-1},x_i,y_{i+1},\dots,y_n) dx_i $$ and repeating this for each coordinate, you conclude (3). Next, let $\alpha$ be a multiindex with $|\alpha|\leq N$, then (2) gives $$ |(D^\alpha\psi)(x)| \leq \max_{y\in Q}|(T^{|\alpha|}\psi)(y)| \leq \max_{y\in Q}|(T^N\psi)(y)|$$ and by (3) you have for each $y\in Q$ $$ |(T^N\psi)(y)| \leq \int_Q |(T^{N+1}\psi)(z)| dz. $$ For (6) there is nothing more to elaborate, as $\int_Q \dots = \int_K\dots $ for $\phi\in\mathcal{D}_K$.
----- (further edit after OP's comments)
If $T\psi=0$, then by (3) $\psi=0$, which is the injectivity of the linear operator $T$. $\Lambda_1$ is a linear functional, and by (8), it is continuous on $Y$ equipped with $L^1$-norm. Then Hahn-Banach says, it can be extended to all of $L^1(K)$, so there is an
element of $L^\infty$ representing $\Lambda_1$ like in (9).
Finally, if a distribution $\Lambda$ has compact support $K$, there are other compact sets $K_1\subset K_2\subset \Omega$ with $K\subset\operatorname{int } K_1\subset \operatorname{int }K_2$. Then the theroem is valid for all $\phi\in\mathcal{D}_{K_2}$. But this is implies the validity of the theorem for all $\phi\in\mathcal{D}(\Omega)$: Let $\chi\in\mathcal{D}$ with $\chi(x)=1$ for $x \in K_1$ and $\operatorname{supp }\chi\subset K_2$. Then for $\phi\in\mathcal{D}(\Omega)$ you have $ \Lambda(\phi) = \Lambda(\chi\phi)$ and for this $\chi\phi$ you have the validity of the theorem.