How does the failure of integral closure of a coordinate ring relate to the cusp/singularity we see in the corresponding variety?

Let $R' = R[y/x]$. We can think of $R \subset R' \subset \operatorname{Frac}(R)$, the field of fractions of $R$.

I claim that $R' \cong \mathbb{C}[t]$. Write $t = y/x \in R'$. In $R'$, $xt=y$, so $x^2 t^2 = y^2 = x^3$. Since $R'$ is an integral domain, $t^2 = x$. And then $t^3 = y$. So (with some hand-waving) $R' = \mathbb{C}[x,y,t]/(y^2-x^3) \cong \mathbb{C}[x,y,t]/(t^2-x,t^3-y) \cong \mathbb{C}[t]$.

If you are familiar with prime spectra of rings, the inclusion $R \subset R'$ gives a map $\operatorname{Spec} R' \to \operatorname{Spec} R$.

If you would rather work with affine varieties, $R = \mathbb{C}[x,y]/(y^2-x^3)$ means $\operatorname{Spec}(R) = V(y^2-x^3)$. What about $R'$? Here $\operatorname{Spec}(R') = V(t^2-x,t^3-y) \subset \mathbb{C}^3$. This is a smooth (nonsingular) curve, parametrized by $t \mapsto (t^2,t^3,t)=(x,y,t)$.

The projection $(x,y,t) \mapsto (x,y)$ gives a map of curves $V(t^2-x,t^3-y) \to V(y^2-x^3)$.

The projection $(x,y,t) \mapsto (x,y)$ has a critical point when the tangent line of the curve is "vertical", i.e., in the kernel of the projection to the $xy$-plane. The derivative of the parametrization is $(2t,3t^2,1)$, which is equal to $(0,0,1)$ when $t=0$. So $t=0$ (or $(x,y,t) = (0,0,0)$) is a critical point of the projection from this curve to the $xy$-plane, and $(x,y)=(0,0)$ is a critical value.