This is an annoying question and I apologize in advance.
Vakil defines a relative proj sheaf of algebras on $X$ to be such that there is a cover of $\operatorname{Spec}(A)$ s.t above each we have $S^A$ a $\mathbb{Z}_{\geq 0}$ graded quasicoherent algebra (this just means each graded component is a quasicoherent module).
Importantly, he requires that $S^A$ at grade $0$ is precisely $\operatorname{Spec}(A)$, and not a quotient (i.e one could fathom a definition that $S^A$ just means something $A$ acts on and doesn't change degree, but this isn't the case).
Given such a sheaf one defines a scheme by taking the proj construction on each of those, with a natural mapping to $X$, so that we get from such a sheaf $F$ on $X$ a map $$\operatorname{Proj}(F) \to X$$
Projective morphisms are then those that arise in this way for finite-type generated in degree $1$ $F$.
My question is then why if $$ Z \to \mathbb{P}^1_X$$ is a closed embedding then it is a projective morphism. Even in the case $X=\operatorname{Spec}(A)$, obviously $Z$ is given by a Proj construction which should finish, but at degree $0$ it will be a quotient of $A$, and not precisely $A$. An example to keep in mind is $X = \operatorname{Spec}(A) = \mathbb{A}^1_k$, and $Z$ is the copy of $\mathbb{P}^1$ over a point.
We can try to solve this by just setting in the graded algebra describing $Z$ the 0th component to $A$, but this doesn't work, it really gives a different scheme.
I think the real solution lies in just instead of requiring $S_0 = A$, instead work with graded over $A$ to mean there is a map $A \to S_0$.
If you want to see this contradicting a specific exercise (or at least the obvious 'solution' to it); though I don't recommand it, since the above baby case holds my concern:
Exercise 17.3.A says suppose $ \pi :X \to Y$ is a morphism, then it is projective if there is a finite type quasicoherent sheaf $F_1$ on $Y$ and a closed embedding $X \to \mathbb{P}(F_1)$ over $Y$ (\mathbb{P}(F_1) means take the graded symmetric algebra that $F_1$ generates and apply the proj construction to it). The obvious solution wants us to say take the graded module of (\mathbb{P}(F_1)) and divide it by the kernel of the map $X \to \mathbb{P}(F_1)$ (i.e $X$ is this divided by some ideal). The problem is that at grade $0$ we're not supposed to be dividing by anything according to Vakil.
Best Answer
The fact that Vakil lets you get away with schemes isomorphic to $\operatorname{Proj} S_\bullet$ for $S$ a finitely generated algebra over $A$ resolves this issue and lets you get past this trouble.
The key thing here is that if $S_\bullet$ is a graded ring and $S_0'$ is any ring with a homomorphism $\varphi:S_0'\to S_0$, the graded ring $S'_\bullet$ defined by $$S'_i=\begin{cases} S_0' & i=0 \\ S_i & i > 0 \end{cases}$$ with multiplication defined by $s'_0\cdot s_d = \varphi(s'_0)\cdot s_d$ for $s'_0\in S'_0$ and $s_d\in S_d$ has $\operatorname{Proj} S_\bullet \cong \operatorname{Proj} S'_\bullet$.
To prove this, we will show that a graded prime ideal $P$ of $S_\bullet$ not containing the irrelevant ideal $S_+$ is determined by its intersection with $S_+$. By the condition that $P$ is a graded ideal not containing $S_+$, there must be some homogeneous element $x\in S_+$ not in $P$. Letting $s_0\in S_0$ be arbitrary, by primality of $P$ we have that $s_0x\in P$ exactly when $s_0\in P$, and so we can recover $P\cap S_0$ from $P\cap S_+$.
Letting $S$ be a graded $A$-algebra where $S_0$ isn't necessarily equal to $A$, we can apply the above construction with $S'_0=A$ and $\varphi:S_0'\to S_0$ the structure map for $S_0$ as an $A$-algebra to see that $\operatorname{Proj} S_\bullet$ is isomorphic to $\operatorname{Proj} S'_\bullet$, which is a graded ring over $A$ in the sense of Vakil.