Is it true that a line bundle is relatively ample iff its restsriction to fibers is? If so, what would be the reference?

If you admit the map to be proper and the schemes to be reasonably good it is true. A reference I know is Lazarsfeld's book "Positivity in algebraic geometry", paragraph 1.7. 


EDIT : my previous answer was wrong. Thanks to BConrad for pointing it out. Here is a counterexample if the map is not proper. Let $X$ be the plane, let $Y$ be the blowup of the plane in one point, and $U$ be $Y$ with one point of the exceptionnal divisor removed. Let $f_U:U\to X$ be the projection and consider the line bundle $\mathcal{O}_U$. Since the fibers of $f_U$ are affine (either points or the affine line), $\mathcal{O}_U$ becomes ample when restricted to fibers of $f_U$. However, $\mathcal{O}_U$ is not $f_U$ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$, but we can compute : $$H^0(U,\mathcal{O}_U^N)=H^0(U,\mathcal{O}_U)=H^0(Y,\mathcal{O}_Y)=H^0(X,\mathcal{O}_X),$$ where the second equality comes from property $S2$ and the third holds because $f_* \mathcal{O}_Y=\mathcal{O}_X$. Hence, $H^0(U,\mathcal{O}_U^N)$ cannot distinguish between two points of the exceptionnal divisor, and $\mathcal{O}_U$ cannot be ample on $U$. Warning : what follows is false. I kept it here so that the comment below remains understandable. Here is a counterexample if the map is not proper. Consider the inclusion $f$ of the plane minus a point $U$ in the plane $X$. The line bundle $\mathcal{O}_U$ is ample restricted to the fibers of $f$ (they're points...). However, it is not $f$ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$. Choosing $N>>0$, we would get $$H^1(U,\mathcal{O}_U)=H^1(U,\mathcal{O}_U^N)=0.$$ But a simple computation via Cech cohomology shows that this cohomology group is not trivial (in fact, infinite). 

