Setup of the problem : Let $f : X \rightarrow Y$ be a flat(smooth) projective morphism with relative dimension $d \geq 1$ and relative very ample sheaf $\mathcal{O}(1)$.
Q : Is it possible to find a section of $\mathcal{O}(n) \otimes f^*\mathcal{L}$ for all $n >> 0$ and some (Can we write ample line bundle $\mathcal{L}$ on $Y$?) line bundle $\mathcal{L}$ on $Y$, such that $Z(s)$, the zero set of $s$, is flat(smooth) over $Y$?
If $Y = k$, and we replace flatness by smooth, we get the standard Bertini theorem. I presume that this is a weaker condition and perhaps it admits some generalization in this form.
My thoughts so far : Let $E$ be a vector bundle over a smooth projective scheme $Y$ over $k(= \overline{k})$, and let $X = \mathbb{P}(E) \xrightarrow{\pi} Y $, the projective bundle associated to $E$. For any $n$ large enough, I can find $s \in \Gamma(X, \mathcal{O}(n) \otimes \mathcal{L})$ for an ample line bundle $\mathcal{L}$ on $Y$. This can be done as follows : We know that $\Gamma(X, \mathcal{O}(n) \otimes \mathcal{L}) = \Gamma(Y, Sym^n E \otimes \mathcal{L})$. For $n$ large enough the rank of $Sym^nE$ is greater than $dim(Y)$ and thus can find a everywhere non-vanishing section of $Sym^n E \otimes \mathcal{L}$ for some ample line bundle $\mathcal{L}$. This section gives us a section of $\mathcal{O}(n) \otimes \mathcal{L}$ such that $dim(Z(s)) < dim(X)$ and $Z(s) \rightarrow Y$ is flat. The issue is we cannot repeat this argument again with $X = Z(s)$, since $X$ maynot be projective bundle of a vector bundle.
Hence the reformulation of the above problem in a more general setup. I am in fact only interested in the case of $Y$ and as a consequence $X$ being projective varieties.
Thanks.
Added on August 16, 2021 : I have been unable to find such a theorem in the literature but I am far from an expert. I would appreciate a proof and if not perhaps an outline or a counterexample in case the desired statement is false for both the smooth and flat scenario.