$$B= \begin{pmatrix} 0 & 1 & 0 \\ -4 & 4 & 0 \\ -2 & 1 & 2 \end{pmatrix}$$ I need to find Jordan decomposition of B.
My sketch:
We find Jordan's matrix first $A$ similar to matrix $B$, and matrix $C \in (3,\mathbb{C})$, such that $B = C^{-1}AC$ We have: $$F_{B}(t)=\text{det}(\left( \begin{array}{ccc} -t & 1 & 0 \\ -4 & 4-t & 0 \\ -2 & 1 & 2-t \end{array} \right))=(2-t)(-4t+t^{2}+4)=(2-t)^{3}$$ $$\text{SP}(B) = {2}$$ $\text{SP}$ means the spectrum of endomorphism. We are now defining a subspace $$\begin{bmatrix} -2 & 1 & 0 \\ -4 & 2 & 0 \\ -2 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \iff x_{1}= -\frac{1}{2}x_{2} $$ So we have two eigenvectors:
$$V_{1}=\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} $$ $$ V_{2}=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ so, $2$ is the dimension of the eigenspace, therefore A is not diagonalizable and Jordan canonical form is
$A=\begin{pmatrix} -2 & 0 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2 \end{pmatrix}{}$
I have issue with last sequence. Why $B$ have this specify form $J$ with 1 for second row?
Im not sure if I understood you question correctly but :
Firstly since the eigenvalue is 2 we have: $A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}{}$ Also by changing the arrangement of the eigenvectors/extend-eigenvectors in the transformation matrix we could also get: $A=\begin{pmatrix} 2 & 1& 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}{}$ hence the 1 being in the second row is not important or fixed.
In general you can change the order of the Jordan-Blocks by as you wish by changing the order of the vectors in the transformation matrix