I’m afraid that this statement is false. Consider the case $X=Y=\mathrm{Spec}\,\mathbb{C}(T)$ and $K=\mathbb{C}(T)$, $f$ is the identity, $R$ is the ring of rational fractions defined at $1$, and $X \rightarrow Y$ is $f \in \mathbb{C}(T) \rightarrow f(T^2)$.
If there were $R’$, $K’$ as claimed, then the intersection of $R’$ and $K(X)$ yields a DVR $R’’ \subset \mathbb{C}(T)$ integral over $R$ (beware the embeddings though). But $R’’$ is normal, so $R’’$ is the integral closure in $\mathbb{C}(T)$ of the ring of fractions $f(T^2)$ with $f(T)$ defined at $1$.
But said integral closure is the ring of fractions defined at $1$ and $-1$, which isn’t local. Indeed, if there are polynomials $a_0,\ldots,a_n$ with $a_n(1) \neq 0$ satisfying $\sum_{k=0}^n{a_k(T^2)f^k(T)}=0$, where $f$ is a rational fraction, then the denominator of $f$ divides $a_n(T^2)$ so vanishes neither at $-1$ nor at $1$.
Here is a counter-example. Consider the blow-up
$$X = \operatorname{Bl}_0 \mathbb A^2 \to \mathbb A^2,$$
and let $E \subset X$ be the exeptional curve. According to [1, Exc. II.4.12 (b.2)], the local ring $R_v = \mathcal O_{X, \varepsilon}$ at the generic point $\varepsilon \in E$ is a valuation ring which dominates the local ring $R = \mathcal O_{\mathbb A^2, 0}$. Let's calculate this!
We have
$$X = \{(x,y, [\xi:\eta]) \,|\, x \eta = y \xi\} \subset \mathbb A^2 \times \mathbb P^1,$$
so the open set $\{\eta \neq 0\}$ is affine with coordinate ring $\mathbb C[y, x/y] \subset \mathbb C(x,y)$. Localizing at the prime ideal $\mathfrak p = (y)$ yields
$$\mathcal O_{X, \varepsilon} = \mathbb C[y, x/y]_{(y)},$$
so we see that the residue field of $\mathcal O_{X, \varepsilon}$ is the function field $\mathbb C(x/y)$, which has transcendence degree $1$ over $\mathbb C$. On the other hand, the residue field of
$$\mathcal O_{\mathbb A^2, 0} = \mathbb C[x,y]_{(x,y)}$$
is just $\mathbb C$, so the two residue fields can never be isomorphic.
For completeness, it is clear that $\mathcal O_{X, \varepsilon}$ is a DVR, because it is a noetherian one-dimensional regular domain. To see that $\mathcal O_{X, \varepsilon}$ dominates $\mathcal O_{\mathbb A^2, 0}$, note that $x = y \cdot \frac{x}{y} \in y \cdot \mathbb C[y, x/y]$. So the inclusion $\mathbb C[x,y]_{(x,y)} \subset \mathbb C[y, x/y]_{(y)}$ maps the maximal ideal $(x,y) \mathcal O_{\mathbb A^2, 0}$ to $y \cdot \mathcal O_{X, \varepsilon}$, and hence $$\mathcal O_{\mathbb A^2, 0} \cap y \cdot \mathcal O_{X, \varepsilon} = (x,y) \mathcal O_{\mathbb A^2, 0}.$$
[1] Hartshorne, Algebraic Geometry
Best Answer
Regarding the first part of your question, that is, giving an example of a valuation in an algebraically closed field, recall that any valuation $\nu$ on a field $K$ has an extension to an overfield $L$. If $L$ is the algebraic closure of $K$, then the value group of $L$ is the divisible closure of the value group of $K$ and the residue field of $L$ is the algebraic closure of the residue field of $K$.
So for example consider the $X$-adic valuation on the power series field $k((X))$ and let us denote it by $\nu$. Then the value group $\nu k((X)) = \mathbb{Z}$ and the residue field $k((X)) \nu = k$. Extending $\nu$ to the algebraic closure $\overline{k((X))}$, we have $\nu \overline{k((X))} = \mathbb{Q}$ and $\overline{k((X))} \nu = \overline{k}$. As a bonus we obtain that when char $k = 0$, then $\overline{k((X))}$ equals the Puiseux series field $P(k)$ if and only if $k = \overline{k}$.
Edit : I do not have a concrete example regarding the second part of your question. However one reason I think we talk of valuations in algebraically closed fields is to make sense of ramification theory of subextensions without ambiguity.