@LeakyNun maybe here's a nicer way of thinking about this. Thinking of $3$ as a function on $\operatorname{Spec} \mathbb{Z}$, you can also think of the map $\mathbb{Z}[x] \rightarrow \mathbb{Z}$ mapping $x\mapsto 3$.

This induces a map $\operatorname{Spec} \mathbb{Z} \rightarrow \operatorname{Spec} \mathbb{Z}[x]$.

If you base change (i.e. tensor) to $\mathbb{F}_p[x]$ downstairs, you get $\mathbb{F}_p$ upstairs, i.e. a map $\operatorname{Spec} \mathbb{F}_p \rightarrow \operatorname{Spec} \mathbb{F}_p[x]$

the intuition here is $\operatorname{Spec} \mathbb{Z}[X]$ is 2-dimensional, so it looks like a surface

you're looking at how this surface fibres over the curve $\operatorname{Spec} \mathbb{Z}$ a curve, - so the vertical lines denote the fibres ("preimage") of the primes $(2), (3), \ldots$ in the map $\operatorname{Spec}\mathbb{Z}[X] \rightarrow \operatorname{Spec}\mathbb{Z}$

the bottom horizontal line is a copy of $\operatorname{Spec}\mathbb{Z}$ in $\operatorname{Spec}\mathbb{Z}[X]$, i.e. the subscheme defined by the ideal $(X)$ (hence you also see the generic point of this guy "on the ri…

oh and just another point on viewing an element of the ring as a function thing

even in the classical AG case, where e.g. you view $x+y$ as associating each maximal ideal $\mathfrak{p}$ in $\mathbb{C}[x,y]$ to something in $\mathbb{C}$,
apriori you're "codomain" is disjoint union of a bunch of $\mathbb{C}$'s

but you would like to identify these $\mathbb{C}$'s together but apriori you can't do that - and really the way to view this, i think, is from what i was saying earlier (as a map to $\mathbb{A}^1_{\mathbb{C}}$