What is an $R$-valued point on a scheme over $R$? An $R$-valued point on an arbitrary scheme $X$ is defined to be a morphism $\mathrm{Spec}(R) \to X$.
(1) What if $X$ itself is a scheme over $R$? In this case, what is the difference between the set of points in $X$ and the set of $R$-valued points? For example, if $R=\mathbb C$ it seems that they agree in some sense, or not?
(2) In general, is there any reasonable structure on the set of all $R$-valued points on a scheme $X$ (i.e. the set of all morphisms from $\mathrm{Spec}(R)$ to $X$)?
 A: Here is a nice fact: if $R$ is a local ring, then an $R$-valued point of a scheme $X$ is uniquely determined by a point $x\in X$, and a local homomorphism $\mathcal{O}_{X,x} \to R$.  So it's just a point with some extra algebraic structure.  If $R$ is a field, these are just embeddings $k(x) \to R$.
I'm not sure how complicated things get when $R$ is a DVR, but it's at least as complicated, since a point over $R$ gives rise to a point over the residue field.
In general, the set of $R$-valued points of a scheme $X$ should be thought of as the image under the Yoneda embedding / functor of points, $Z \mapsto \operatorname{Hom}(Z,X)$, when $Z=\operatorname{Spec}(R)$ is affine.  It inherits various bits of structure (e.g. pullbacks) from this embedding.
By the way: it's common to define all of this over some base ring/scheme, rather than working implicitly over $\mathbb{Z}$.  This makes the math cleaner.  For example, if we're working over $\mathbb{Z}$, there is a $\mathbb{C}$-point of $\operatorname{Spec}(\mathbb{C})$ for every automorphism of $\mathbb{C}$.  However, if we require everything to be a $\mathbb{C}$-morphism, then $\operatorname{Spec}(\mathbb{C})$ has only one $\mathbb{C}$-point.  If we require everything to be an $\mathbb{R}$-morphism instead, then $\operatorname{Spec}(\mathbb{C})$ has two $\mathbb{C}$-points, one arising from conjugation.
