For a linear map $f: V \to V$ if $f^2$ is diagonalizable and $\ker f = \ker f^2$ then is $f$ diagonalizable?

The answer is no.

Counter-example: let $\mathbb F$ be the algebraic closure of $\mathbb F_2$, $n=2$ and $f$ have matrix representation

$A= \displaystyle \left[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right]$

Then $f$ is an involution but not diagonalizable.

Now when $\text{char } \mathbb F \neq 2$ the the claim is true since the eigenvalue 0 is semi-simple and when we restrict to the subspace on which $f$ is invertible, and reference e.g. my triangularization argument here: $A \in M_n(\mathbb{C})$ invertible and $A^2$ is diagonalizable. Prove $A$ is diagonalizable we see $f$ must be diagonalizable on this subspace.