Intuition on base change of schemes

First paragraph: yes. Second paragraph: getting equations is going to be gross in general, although it depends on the nature of $S'$ and $S$. Third paragraph: no.

Maybe this will be a motivating example. Let $S$ be Spec of an algebraically closed field, and let $S'$ be Spec of a much larger algebraically closed field (e.g. adjoin a transcendental to your field, then algebraically close that). If you take an $S'$ scheme and "think of it as an $S$-scheme" you will get a monstrosity that's not of finite type (e.g. in the last paragraph you are asserting that $\mathbb{C}[t]$ is isomorphic to $\overline{\mathbb{Q}}[t]$ as a $\overline{\mathbb{Q}}$-algebra).

The lesson to take away is that base change is the reasonable, good functor, and the "forgetful functor" is the ugly one, the opposite of the way you'd think it would be on first glance.

Just to add to the excellent answer of countinghaus, regarding your first question, your morphism $f$ is going the wrong way. You want a morphism $f:S^\prime\rightarrow S$ to talk about $X\times_SS^\prime$, not $S\rightarrow S^\prime$. Anyway, when $X$ is the spectrum of an $A$-algebra $R$, on choosing generators, and assuming this algebra is finitely presented, it is isomorphic to $A[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$ for some $n\geq 0$ and some $f_j\in A[x_1,\ldots,x_n]$. The base change is the spectrum of $R\otimes_AB$, tensor product with respect to the ring map $\psi:A\rightarrow B$ corresponding to $S^\prime\rightarrow S$. The canonical $B$-algebra map

$R\otimes_AB=A[x_1,\ldots,x_n]/(f_1,\ldots,f_m)\otimes_AB \rightarrow B[x_1,\ldots,x_n]/(\psi(f_1),\ldots,\psi(f_m))$

is an isomorphism, so indeed, the base change is "cut out" by the polynomials $f_j$, viewed via $\psi$ as having coefficients in $B$.