You can reverse engineer a zillion examples by starting with the normalization: Let $S$ be a normal local domain. Let $I$ be any nonzero ideal of $S$ and let $A$ be a subring of $S/I$. Take $R = \{ f \in S : [f] \in A \}$, where $[ \ ]$ denotes reduction modulo $I$. Then $\mathrm{Frac}(R) = \mathrm{Frac}(S)$, since $I \subset R$, and it is easy to see that $S$ is the normalization of $R$ in $\mathrm{Frac}(S)$.

To get the cusp, take $S = k[t]$, $I = (t^2)$ and $A = k \subset S/I = k[t]/t^2$.

You can reverse engineer a zillion examples by starting with the normalization: Let $S$ be a normal local domain. Let $I$ be an ideal of $S$ with $\sqrt{I}=\mathfrak{m}$ and let $A$ be a subring of $S/I$. Take $R = \{ f \in S : [f] \in A \}$, where $[ \ ]$ denotes reduction modulo $I$. Then $\mathrm{Frac}(R) = \mathrm{Frac}(S)$, since $I \subset R$. I claim that $S$ is the normalization of $R$. Clearly, $S$ is a normal subring of $\mathrm{Frac}(R)$ containing $R$. Using that $\sqrt{I}= \mathfrak{m}$, one can show that $S$ is a finite $R$-module, so $S$ integral over $R$. (An earlier version of this answer had a weaker condition on $I$, but then $S$ need not be integral over $R$. For example, take $S=k[x,y]$, $I=(x)$ and $A = k \subset k[x,y]/(x)$. Then $y$ isn't integral over the (nonnoetherian) ring $R$.)

To get the cusp, take $S = k[t]$, $I = (t^2)$ and $A = k \subset S/I = k[t]/t^2$.