Calculating generalized eigenvectors

The eigenspace for the eigenvalue $\lambda=0$ is given by: $$ A\mathbf x= \begin{bmatrix}0&1&-1&-1\\ 0&0&0&0\\ 0&-1&2&2\\ 0&1&-2&-2 \end{bmatrix} \begin{bmatrix} x\\y\\z\\t \end{bmatrix}= \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix} $$ that gives: $$ \begin{bmatrix} x\\y\\z\\t \end{bmatrix}= \begin{bmatrix} x\\0\\-t\\t \end{bmatrix} $$ so we can chose two linearly independent eigenvectors as: $$\mathbf v_1= \begin{bmatrix} 0\\0\\-1\\1 \end{bmatrix}\qquad \mathbf v_2= \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} $$ Now using $\mathbf v_1$ we can find a generalized eigenvector searching a solution of: $$ \begin{bmatrix}0&1&-1&-1\\ 0&0&0&0\\ 0&-1&2&2\\ 0&1&-2&-2 \end{bmatrix} \begin{bmatrix} x\\y\\z\\t \end{bmatrix}=\begin{bmatrix} 0\\0\\-1\\1 \end{bmatrix} $$ that gives a vector of the form $$ \begin{bmatrix} x\\y\\z\\t \end{bmatrix}= \begin{bmatrix} x\\-1\\-1-t\\t \end{bmatrix} $$ and, for $x=t=0$ we can chose the vector $\mathbf w_1=[0,-1,-1,0]^T$

In the same way we can find the generalized eigenvector $\mathbf w_2=[0,2,1,0]$ as a solution of $A\mathbf x=\mathbf v_2$.

Now we have the matrix $$ M=[\mathbf v_1,\mathbf w_1,\mathbf v_2, \mathbf w_2]= \begin{bmatrix} 0&0&1&0\\ 0&-1&0&2\\ -1&-1&0&1\\ 1&0&0&0 \end{bmatrix} $$ with the inverse: $$ M^{-1}=\begin{bmatrix} 0&0&0&1\\ 0&1&-2&-2\\ 1&0&0&0\\ 0&1&-1&-1 \end{bmatrix} $$ and a Jordan decomposition of The matrix $A$ is: $$A= \begin{bmatrix}0&1&-1&-1\\ 0&0&0&0\\ 0&-1&2&2\\ 0&1&-2&-2 \end{bmatrix}= \begin{bmatrix} 0&0&1&0\\ 0&-1&0&2\\ -1&-1&0&1\\ 1&0&0&0 \end{bmatrix} \begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0&1\\ 0&1&-2&-2\\ 1&0&0&0\\ 0&1&-1&-1 \end{bmatrix}= MJM^{-1} $$

This decomposition is not unique, In the sense that the matrix $M$ ( and $M^{-1}$) can be different, because we can chose different eigenvectors and generalized eigenvectors.

If, as in OP, we chose: $$\mathbf v'_1= \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}\qquad \mathbf v'_2= \begin{bmatrix} 1\\0\\1\\-2 \end{bmatrix} $$ than the generalized eigenvectors that satisfies the equations: $$ A\mathbf w'_1=\mathbf v'_1 \qquad A\mathbf w'_2=\mathbf v'_2 $$ becomes: $$ \mathbf w'_1= \begin{bmatrix} 1\\2\\0\\1 \end{bmatrix} \qquad \mathbf w'_2= \begin{bmatrix} 1\\3\\1\\1 \end{bmatrix} $$

(this seems the mistake in OP) and we have a matrix $$ S= \begin{bmatrix} 1&1&1&1\\ 0&2&0&3\\ 0&0&1&1\\ 0&1&-1&1 \end{bmatrix} $$ and a Jordan decomposition: $$ SJS^{-1}=\begin{bmatrix} 1&1&1&1\\ 0&2&0&3\\ 0&0&1&1\\ 0&1&-1&1 \end{bmatrix} \begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} 1&-2&2&3\\ 0&2&-3&-3\\ 0&1&-1&-2\\ 0&-1&2&2 \end{bmatrix}= \begin{bmatrix}0&1&-1&-1\\ 0&0&0&0\\ 0&-1&2&2\\ 0&1&-2&-2 \end{bmatrix}=A $$

Finally, note that the eigenspace of the eigenvalue $\lambda=0$ is the kernel of $A$ and also $\mathbf u_1=A\mathbf e_2$ and $\mathbf u_2=A\mathbf e_3$ are vectors of the kernel, so they are eigenvectors of $A$, and $\mathbf e_2$ and $\mathbf e_3$ are the corresponding generalized eigenvectors, so another matrix that gives a Jordan decomposition is $N=[\mathbf u_1,\mathbf e_2,\mathbf u_2,\mathbf e_3]$