I know that for every symmetric form $f: U \times U \rightarrow \mathbb{K}$, char$\mathbb{K} \neq 2$ there exists a basis for which $f$'s matrix is diagonal.
Could you tell me what happens if we omit assumption about $f$ being symmetric?
Could you give me an example of non symmetric bilinear form $f$ which cannot be diagonalized even if we change the basis of one the spaces?
Note that for a bilinear form, the so-called "diagonalisation" is not diagonalisation via similarity, but diagonalisation via congruence. That is, if $A$ is the matrix for $f$ w.r.t. some basis, we look for an invertible matrix $P$ such that $P^TAP$ is equal to some diagonal matrix $D$. Yet, if $P^TAP=D$, then $A=(P^{-1})^TDP^{-1}$ and hence $A$ is necessarily symmetric.
(Put it another way, if $f$ is diagonalisable, then it's matrix w.r.t. some basis is diagonal and hence symmetric. Therefore $f$ is symmetric.)