The actual statement holds in a much more general setting.
Let $\Phi:V\to V$ be any mapping, and let $f:V\to K$ be a function on $V$. Then we get a new function $\Phi^\ast f:V\to K$ defined by $v\mapsto f(\Phi(v))$.
In your case, $V$ is the space of kets, and $\Phi$ is a linear operator on it. A linear map $f:V\to\mathbb C$ is a bra. (Let's stay in the finite dimensional case to not have to worry about continuity and so.) Since $\Phi$ is linear, it is not hard to see that if $f$ is linear, then so is $\Phi^\ast f$. That is all there really is about how $\Phi$ acts on bra's.
If you take this further, then you see how the matrix of $\Phi$ on kets transforms when expressed in the dual basis of bra's. This is elaborated in the other answers.