First, a small correction: you say something about $X$, $Y$ and $X\times_Y \operatorname{Spec} k(y)$ being of finite type here, but you really should be talking about $X\to\operatorname{Spec} k$, $Y\to\operatorname{Spec} k$, and $X\times_Y \operatorname{Spec} k(y)\to \operatorname{Spec} k(y)$ being of finite type, because being of finite type is a property of morphisms. When one talks about a scheme having a property of a morphism of schemes like this, it is usually assumed that what one means is the canonical morphism to $\operatorname{Spec} \Bbb Z$ has this property. This is problematic for you because no $\Bbb C$-scheme can be of finite type over $\Bbb Z$ for cardinality reasons, for instance. You also make a conclusion about the finite-type-ness of some scheme based on it being a point, but this is inappropriate: $\operatorname{Spec} k[x_1,\cdots]/(x_1,\cdots)^2$ is a single point, but not finite type over $\operatorname{Spec} k$, for example. Basically, don't forget your base!
Let's remember the definition of a finite type morphism: a morphism of schemes $f:X\to Y$ is called finite type if it's quasi-compact and locally of finite type. Quasi-compact means that the inverse image of a quasi-compact set is again quasi-compact, and locally of finite type means that if we have any two open affine schemes $\operatorname{Spec} A\subset X$ and $\operatorname{Spec} R\subset Y$ with $f(\operatorname{Spec} A)\subset \operatorname{Spec} R$, then the induced map on rings $R\to A$ makes $A$ a finite-type $R$-algebra. (We say a ring map $R\to A$ is of finite type if $A$ is isomorphic to a quotient of $R[x_1,\cdots,x_n]$ as an $R$-algebra.)
We'll deal with being locally of finite type first. To be specific:
Lemma (ref). Suppose $X\to Y$ is a morphism of schemes over some base $S$. If $X$ is locally of finite type over $S$, then $X\to Y$ is locally of finite type.
Proof. The condition on rings is equivalent to asking that if $A\to B \to C$ is a sequence of ring maps so that $C$ is finitely generated over $A$, then it's finitely generated over $B$. This is straightforwards: write $C=A[x_1,\cdots,x_n]/J$ and suppose $B$ is generated as an $A$-algebra by some collection of elements $\{y_\alpha\}_{\alpha\in A}$. Let $\overline{y_\alpha}$ denote the image of $y_\alpha$ in $C$. Now I claim that $B[x_1,\cdots,x_n]/(J,y_\alpha-\overline{y_\alpha})\cong C$, where I mean the ideal generated by the images of all elements of $J$ in $B$ and all elements of the form $y_\alpha-\overline{y_\alpha}$ as $\alpha$ ranges over the index set $A$. $\blacksquare$
This previuous lemma is totally general, which is nice! On the other hand, it is not true in general that if $X\to Y$ is a morphism of schemes over a base $S$ and $X\to S$, $Y\to S$ are quasi-compact then one has $X\to Y$ quasi-compact. Examples of this necessarily involve the failure of $Y\to S$ to be quasi-separated, which is probably something you won't see in nature for a while if you're a newer algebraic geometer. (Such an example is necessarily a non-Noetherian scheme, for instance, so if you aren't venturing out of the garden of Noetherian schemes, you're fine.)
In our case where we work over a field, we may conclude the proof as follows. Since $X$ is finite type over a field, it's a noetherian topological space, so every subset of it is quasicompact. This implies that every morphism out of $X$ is quasicompact: the preimage of any set under any morphism coming out of $X$ will be quasicompact, so the definition of a quasicompact morphism is trivially satisfied. Thus, if $X$ and $Y$ are schemes of finite type over a field, then any morphism $X\to Y$ is also of finite type.
In EGA IV part three we find the following result:
Corollary 9.6.4: Let $f:X \to S$ be a proper and finitely presented morphisms of schemes and $\mathcal{L}$ a line bundle on $X$. Then the set $U \subset S$ of $s \in S$ such that $\mathcal{L}_s$ is relative ample for $f_s$ is open in $S$, and the restriction of $\mathcal{L}$ to $f^{-1}(U)$ is relative ample for $f:f^{-1}(U) \to U$.
This answers your second question because $\mathcal{L}_s$ being relatively ample for $f_s$ is equivalent to $\mathcal{L}_s$ being ample on $X_s$. Your first question follows by taking an affine open $V \subset U$ and applying part 3 of https://stacks.math.columbia.edu/tag/01VJ.
Best Answer
If $\DeclareMathOperator{\Spec}{Spec} f: X:=\Spec(B) \rightarrow S:=\Spec(A)$ and if $L\in Pic(A)$ it follows $\DeclareMathOperator{\Sym}{Sym}\mathbb{V}(L^*):=\Spec(\Sym_A^*(L^*))$, hence there are isomorphisms
$$\mathbb{V}((f^*L)^*):=\Spec(\Sym_B^*((B\otimes_A L)^*) \cong \Spec(B\otimes_A \Sym_A^*(L^*)) \cong X\times_S \mathbb{V}(L^*).$$
Note For a locally trivial $A$-module $L\DeclareMathOperator{\Hom}{Hom}$ it follows $\Hom_B(B\otimes_A L, B) \cong B\otimes_A \Hom_A(L,A)$, hence
$$(B\otimes_A L)^* \cong B\otimes_A L^*.$$
Moreover
$$ \Sym_B^*((B\otimes_A L)^*) \cong B\otimes_A \Sym_A^*(L^*).$$