The answer to the first question is yes, $F$ can be expressed as a semidirect prodcut in many different ways. Following i.m.solovelchik's comment, let $f$ be any epimorphism from $F$ to a free group, and let $K:=\ker f$. For example, if $a$ is one of the free generators, then we could define $f$ by $f(x) = a$ for all members of a free generating set containing $a$.
Then $F/K$ is free, and so $K$ has a complement $H$ in $F$ and $F = K \rtimes H$.
But the answer to the second question is no, we cannot write $F = K \rtimes H$ with both $K$ and $H$ free of finite rank. It can be shown that any nontrivial normal subgroup of $F$ of infinite index in $F$ has infinite rank, and so if $H$ and $K$ are both nontrivial, then $K$ must have infinite rank.