Misconception on Jordan Canonical Form

$\DeclareMathOperator\ker{ker}\DeclareMathOperator\id{id}$Here's the idea: Denote by $H_i=\ker(f-\lambda\id)^i$ the generalized eigenspaces and note that they form a chain $H_1\subseteq H_2 \subseteq \cdots$. Pick a vector $v_k\in H_k\setminus H_{k-1}$. Now define the vectors $v_{k-1},\dots,v_1$ by letting $v_i = (f-\lambda\id)(v_{i+1})$ for $i=k-1,\dots,1$. Note that $v_i\in H_i\setminus H_{i-1}$. Hence, the vectors $v_1,\dots,v_k$ will be linearly independent and by construction we have \begin{align*} f(v_k) &= \lambda v_k + v_{k-1}, \\ f(v_{k-1}) &= \lambda v_{k-1} + v_{k-2}, \\ &\,\,\,\vdots\\ f(v_2) &= \lambda v_2 + v_1, \\ f(v_1) &= \lambda v_1. \end{align*} Thus, the subspace $U=\langle v_1,\dots,v_k\rangle$ is $f$-invariant and the matrix of $f$ on $U$ with respect to the basis $v_1,\dots,v_k$ is $$ \begin{pmatrix} \lambda & 1\\ &\lambda & 1 \\ & & \lambda \\ & & & \ddots & 1 \\ & & & & \lambda \end{pmatrix}. $$

$\DeclareMathOperator{\Id}{Id}$In Your case You already found a vector $v\in ker (M-2\Id)^2\backslash ker(M-2\Id)$, namely $v=(0,0,1,0)^T$ and $(M-2\Id)v=(1,4,0,0)^T$. Furthermore $M(1,4,0,0)^T=2(1,4,0,0)^T$ and $Mv=2v+(1,4,0,0)^T$. This corresponds to the Jordan block $$\begin{pmatrix}2&1\\0&2\end{pmatrix}.$$ The subspace $\operatorname{span}\{(0,0,1,0)^T,(1,4,0,0)^T\}$ is thus an invariant subspace.