Given a vector space $V$ of dimension $4$ and a base $\{v_1,v_2,v_3,v_4\}$, let $f$ be an endomorphism of $V$ such that $f^3=0$ and moreover $f(v_1)=f(v_2)=v_3$, $f(v_3)=kv_4$, and $f(v_4)\in\left<v_1,v_4\right>$, where $k$ is a real parameter.
I should find the characteristic polynomial $\chi_f$, as well as the minimal one $m_f$, and the Jordan normal form of $f$, depending on $k$ and $f(v_4)$.
Now, here's my thoughts: $f^3=0$ means that either $m_f(t)=t^2$ or $m_f=t^3$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=k\cdot f(v_4) $$so I guess one should consider separately $k=0$, which lets $f(v_4)$ be a parameter, and $k\ne0$ which forces $f(v_4)=0$.
How should I continue?
Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $\chi_f(\lambda)=\lambda^4$.
Assume first that $k\ne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3\times 3$ Jordan block. So the Jordan form of $f$ is $$ \begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}. $$ Take now the case $k=0$. If $b\ne0$, then $$ x=\tfrac ab\,v_1+\tfrac a{b^2}\,v_3+v_4 $$ satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $a\ne0$, then $v_3,v_1,\tfrac1a\,v_4$ is a basis for a $3\times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2\times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.
In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.
If $k=0$ and $a=0$, the Jordan form is $$ \begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}. $$
If $k=0$ and $a\ne0$, or if $k\ne0$, the Jordan form is $$ \begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}. $$