For smooth projective varieties $Y \subset \mathbb{P}^n$ we know that the coordinate ring $S(Y)$ may or may not be integrally closed. If $S(Y)$ is normal, we say $Y$ is projectievly normal. This notion measures normality of the affine cone $C(Y) \subset \mathbb{A}^{n+1}$ over $Y$.
A standard example of non-projectively normal variety is the twisted quartic in $\mathbb{P}^1 \hookrightarrow \mathbb{P}^3$, given by the embedding $$ [u:v]\mapsto [u^4:u^3v:uv^3:v^4] $$ This embedding is obtained via projecting the rational normal curve in $\mathbb{P}^1 \hookrightarrow \mathbb{P}^4$ given by $$ [u:v]\mapsto [u^4:u^3v:u^2v^2:uv^3:v^4] $$ on a suitable hyperplane in $\mathbb{P}^4$.
Call $Y_1$ the twisted quartic in $\mathbb{P}^3$ and $Y_2$ the rational normal curve in $\mathbb{P}^4$. We have that $S(Y_2)$ is the integral closure of $S(Y_1)$. Therefore, $C(Y_2)$ is the normalization of $C(Y_1)$.
Question
My question is the following. How do I visualize the map $C(Y_2) \rightarrow C(Y_1)$?
Edit
Using the equations in this question, taking the jacobian matrix and checking for where the rank drops below 2, it looks like the cone $C(Y_1)$ is smooth away from the vertex. Thus, $C(Y_1)$ is already $R_1$. Is then the normalization happening just at level of structure sheaves, to ensure the $S_2$ property?