(Supposing the Jordan representation exists) Since a Jordan block associated to $\lambda$ has the form $J= \lambda I + N$, where $I$ is the identity matrix and $N$ a nilpotent matrix. One could find a base where the nilpotent operator $f$ of order $r$ has the matrix representation of $N$, which in turn would be a block diagonal matrix with some Jordan blocks of dimension at most $r$ (with at least one block of dimension equal $r$), all associated to $\lambda=0$.
However, this form always exist if we are working on a algebraically closed field. What happens if the vector field $V$ is not algebraically closed? Can I still say that there is a Jordan representation of $f$? I was thinking maybe there is a way of showing that $\det(f)=\lambda^{\dim(V)}$.
I'm assuming the vector space is final dimensional.
A linear operator $T$ has a Jordan canonical form (there is a Jordan canonical basis for $V$ relative to $T$) if and only the characteristic polynomial of $T$ splits into linear factors; when the ground field is algebraically closed you get that "for free", but even over non-algebraically closed fields some operators will have Jordan canonical forms (even if they are not diagonalizable).
A nilpotent operator $N$ always satisfies the polynomial $x^k$ for some $k\gt 0$. That means the minimal polynomial is a power of $x$. Since every irreducible factor of the characteristic polynomial must divide the minimal polynomial, that means the characteristic polynomial of $N$ must be $(-1)^nx^n$, where $n$ is the dimension of $V$. Thus, the characteristic polynomial of $N$ splits into linear factors, so $N$ will necessarily have a Jordan canonical basis. This holds over any ground field, algebraically closed or not.