Let $S$ be a Noetherian scheme, $f : X\to S$ a smooth quasi-projective morphism, $g : X\to Y$ a morphism of finite type, and $h : Y\to S$ a smooth projective morphism with $h\circ g =f$.
Can we say anything about the scheme-theoretic image of $g$?
For example, let $Z$ be the scheme theoretic image of $g$, with $i : Z\to Y$ the closed immersion, and $p : Z\to S$ the structure morphism $h\circ i$.
Is $p$ an lci morphism?
It's very hard to say anything. In fact, if $S = \operatorname{Spec} \mathbf C$, then any (integral) projective $k$-variety $Z$ arises in this way. Indeed, since $Z$ is projective, there exists a closed immersion $i \colon Z \hookrightarrow \mathbf P^n$ for some $n$, so we may take $Y = \mathbf P^n$. Then we may choose a resolution of singularities $X \twoheadrightarrow Z$, so that the composition $X \twoheadrightarrow Z \hookrightarrow \mathbf P^n$ is a morphism of smooth projective $S$-schemes whose scheme-theoretic image is $Z$.
For a concrete example to see that $Z$ need not be a local complete intersection, even without using resolution of singularities, we can choose $Y = \mathbf P^2 \times \mathbf P^2$ and $X = \mathbf P^2 \amalg \mathbf P^2$, mapping the first component to $\mathbf P^2 \times p$ and the second to $p \times \mathbf P^2$ for some $p \in \mathbf P^2(\mathbf C)$. The scheme-theoretic image is the union of two planes meeting in a point, which is not Cohen–Macaulay by Hartshorne's connectedness theorem (see e.g. Eisenbud, Thm. 18.12 and Figure 18.2); in particular it cannot be a local complete intersection.
(Even worse, the map $X \to Z$ need not be surjective, e.g. we could even take the same $Z = (\mathbf P^2 \times p) \cup (p \times \mathbf P^2)$ above and take $X = Z \setminus (p,p)$ which is smooth.)
