Consider a section $\mathrm{Spec} R \to \mathcal{C}$. The inclusion of $R$-algebras $R \hookrightarrow K$ induces a canonical morphism $\mathrm{Spec} K \to \mathrm{Spec} R$. Thus we obtain a point $\mathrm{Spec} K \to \mathcal{C}$ (factoring through our section from above). The generic fiber is just the base change along this morphism $\mathrm{Spec} K \to \mathrm{Spec} R$, so $K$-points on $\mathcal{C}$ are the same as $K$-points on $C$.
In simpler terms if $(x_0,...,x_n)$ is a point on $C$ and $x_0, ..., x_n \in R$ then viewing them as elements of $K$ gives a point on $C$.
The harder thing here is the converse (going from a $K$-point to an $R$-point) which is true when $C$ is proper (the valuative criterion allows us to do the scheme version of "clearing denominators").