Let $ F \to E \stackrel{\pi}{\to} B$ be a smooth fibre bundle, so that $F$, $E$, and $B$ are smooth manifolds.
I'm interested in what one can say about the anti-canonical bundle $K_E^*$ of the total space $E$, given the anti-canonical bundles of the fibre $F$ and base $B$. In particular, I want to show that if $K_F^*$ and $K_B^*$ are free (i.e. if their linear systems have no base loci), then $K_E^*$ is free too.
According to this page, the tangent bundle of $E$ splits as $$ TE \cong \pi^*(TB) \oplus T_\pi E \,, $$ where $T_\pi E$ consists of those tangent vectors tangent to the fibres of $\pi$. Taking the determinant bundle of both sides, and noting the determinant commutes with the pull-back, we have $$ K_E^* \cong \pi^*(K_B^*) \otimes \mathrm{det}(T_\pi E) \,. $$ Since the pull-back of a free bundle is free, I know the first factor is free. Also, the tensor product of free bundles is free, so if the second factor is free then $K_E^*$ is too, so we would be finished.
However I'm not quite sure how to deal with $\mathrm{det}(T_\pi E)$, and how it is related to $K_F^*$. If the bundle were trivial, $E = F \times B$, with projection $p: E \to F$, then it must be that $T_\pi E \cong p^*(TF)$, so that $K_E^* \cong \pi^*(K_B^*) \otimes p^*(K_F^*)$, which is free. But I don't know how to treat the general case.
What you want to show is not true.
Maybe the simplest counterexample is the Hirzebruch surface $\mathbf F_3$, meaning the projective bundle $\mathbf P(O \oplus O(3))$ over $\mathbf P^1$.
This is a bundle in which the base and fibres are both $\mathbf P^1$, so the anticanonical bundle of both is very ample, in particular basepoint free.
However, this bundle contains a curve $C$ isomorphic to $\mathbf P^1$ with selfintersection $C^2=−3$. Adjunction then tells you that $K_{\mathbf F_3}^* \cdot C<0$, so $C$ is in the base locus of $K^∗$.