$X$ is an affine scheme of finite-type over $R$ . What is an $R$ valued point of $X$? Usually if $X$ is an affine scheme then an $R$ - valued point is defined by a morphism $\operatorname{Spec}(R)\rightarrow X$. But if the scheme $X$ is over $R$ and is of finite-type, do we get any special structure as we also have a morphism $X \rightarrow \operatorname{Spec}(R)$?
 A: If we require that the morphism picking out the $R$-point is an $R$-morphism (this means that the composition $\operatorname{Spec} R \to X\to \operatorname{Spec} R$ is the identity, as mentioned by Angina Seng in the comments), then we can view the set of such points as a subset of the $R$-points of $\Bbb A^n_R$, which can let us pull back interesting structure on $\Bbb A^n_R(R)$ to our $R$-scheme $X$.
For instance, when $R=\Bbb C,\Bbb R$ or $\Bbb Q_p$, then this recovers the analytic topology on the $R$-points of a variety over $R$, which can be used to apply tools from analysis over these fields to algebro-geometric problems. The complex-analytic topology, GAGA, and things of that nature all benefit greatly from this observation.
Let me now argue that requiring us to take $R$-morphisms is usually the right thing to do. If we don't require that the map picking out our point is an $R$-morphism, then we can get some really wacky things happening. Consider $\overline{\Bbb F_p}$: the Galois group of this over $\Bbb F_p$ is $\widehat{\Bbb Z}$, which is uncountable. So we get uncountably many distinct homomorphisms $\overline{\Bbb F_p}\to \overline{\Bbb F_p}$ which induce uncountably many distinct morphisms of schemes between their spectra, which is much different from the single morphism you get from enforcing that your map must be a $\overline{\Bbb F_p}$-algebra morphism. People generally much prefer thinking about the second scenario to the first, though there are times when the first is appropriate or necessary.
