Let $X$ be a scheme over a field $F$ where $F$ is either the real or the complex numbers.
If $q:Y\to X$ is an algebraic vector bundle over $X$, the $F$-points $q(F):Y(F)\to X(F)$ constitute a (topological) $F$-vector bundle over the (topological) space $X(F)$.
Hence, there are two maps (actually homomorphisms of groups, if I am not mistaken. Perhaps also homomorphisms of rings with respect to the $\otimes$ struncture on both sides. Is this true?) $$ r:K_0(X)\to KO_0(X(\mathbb{R})) $$ and $$ c:K_0(X)\to K_0(X(\mathbb{C})) $$ from the Grothendieck groups of vector bundles.
What can be said about injectivity/surjectivity of $r$ and $c$?
What can be said about inj/surj of $r$ and $c$ for the special case that $X=Spec(R)$ is affine?
These are ring homomorphisms, since adding or tensoring algebraic vector bundles is the same as adding or tensoring the underlying topological bundles. Regarding surjectivity, one is asking something along the lines of whether every topological bundle can be algebraized. The answer to this is no, since even for line bundles there are obstructions to algebraizing a topological line bundle, coming from the Lefschetz $(1,1)$ theorem.
More precisely, if $X$ is smooth and projective, then $K_0(X)_{\mathbb Q}$ (the subscript means "tensor with $\mathbb Q$") is isomorphic to the direct sum of the Chow groups $CH^i(X)_{\mathbb Q}$ (again tensored with $\mathbb Q$), the isomorphism being given by Chern classes (more precisely, by the Chern character).
There are cycle class maps from $CH^i(X)$ to $H^{2i}(X(\mathbb C))$, and there is also a Chern character map from $K_0(X(\mathbb C))_{\mathbb Q}$ to $\bigoplus_i H^{2i}(X(\mathbb C),\mathbb Q)$. The obvious diagram commutes, and so we see that $c$ is neither surjective nor injective in general, because:
By Atiyah--Hirzebruch, the Chern character from $K_0(X(\mathbb C))_{\mathbb Q}$ to $\bigoplus_i H^{2i}(X(\mathbb C),\mathbb Q)$ is an isomorphism.
The cycle class maps $CH^i(X)_{\mathbb Q} \to H^{2i}(X(\mathbb C),\mathbb Q)$ are typically neither surjective nor injective. E.g. when $i = 1$, we are just looking at the divisor class map from $Pic(X)$ to $H^2$ (tensored with $\mathbb Q$), and the kernel is the connected component of $Pic(X)$ (tensored with $\mathbb Q$), which is non-zero if $H^1(X,\mathcal O) \neq 0$, while the image is the Neron-Severi lattice of $H^2$ (again, tensored with $\mathbb Q$), which by the Lefschetz $(1,1)$ theorem is the intersection of the rational cohomology with the $(1,1)$-part of the complex cohomology, and so is a proper subspace of $H^2$ provided that $H^2(X,\mathcal O) \neq 0$.
I haven't thought carefully about the affine case, but I suspect it can be similarly complicated.