$A=\begin{pmatrix} -6 & 0 & 1 & -5 \\ 5 & - 1 & -1 & 5 \\ 0 & 0 & -1 & 0 \\ 5 & 0 & -1 & 4 \end{pmatrix}$
I found the Jordan normal form to be
$J=\begin{pmatrix} -1 & 1 &0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$.
I know next that I have to solve $(A+1)v_2=v_1$ to find $v_1= \begin{pmatrix} 1 \\ -1\\0\\-1 \end{pmatrix}$ and $v_2=\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$, where you can then go on to find two more vectors, but I dont understand why this proccess works, can someone explain? Thank you!
Let's follow the algorithm described here.
It seems that you have found that the characteristic polynomial of $A$ is $$ \chi_A(t)=(t+1)^4 $$ so the only eigenvalue of $A$ is $\lambda=-1$ with algebraic multiplicity $m=4$. Note that \begin{align*} A+I &= \left[\begin{array}{rrrr} -5 & 0 & 1 & -5 \\ 5 & 0 & -1 & 5 \\ 0 & 0 & 0 & 0 \\ 5 & 0 & -1 & 5 \end{array}\right] & (A+I)^2 &= \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{align*} so \begin{align*} \dim\operatorname{Null}(A+I) &= 3 & \dim\operatorname{Null}((A+I)^2) &= 4 \end{align*} we then compute the numbers \begin{align*} d_1 &= \dim\operatorname{Null}(A+I) & d_2 &= \dim\operatorname{Null}((A+I)^2)-\dim\operatorname{Null}(A+I) \\ &= 3 & &= 4-3 \\ & & &=1 \end{align*} so we must fill the boxes $$ \begin{array}{ccc} \Box & \Box & \Box \\ \Box \end{array} $$ with vectors. Note that $e_1=\langle1, 0, 0, 0\rangle$ satisfies $e_1\in\operatorname{Null}((A+I)^2)$ but $e_1\notin\operatorname{Null}(A+I)$. Put $$ v=(A+I)e_1=\left\langle-5,\,5,\,0,\,5\right\rangle $$ so the diagram takes the form $$ \begin{array}{ccc} v & \Box & \Box \\ e_1 \end{array} $$ Since $w=\left\langle1,\,0,\,0,\,-1\right\rangle$ and $x=\left\langle0,\,0,\,5,\,1\right\rangle$ are two linearly independent vectors in $\operatorname{Null}(A+I)$ that yield a basis $\{v, w, x\}$ of $\operatorname{Null}(A+I)$ we may complete our diagram with $$ \begin{array}{ccc} v & w & x \\ e_1 \end{array} $$ This gives $A=PJP^{-1}$ where $$ P = \left[\begin{array}{rrrr} -5 & 1 & 1 & 0 \\ 5 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 \\ 5 & 0 & -1 & 1 \end{array}\right] $$