I don't think this can be done.

It seems to me that your $\widetilde U$ would be a submanifold of $\mathbb C^N$, so it should be a local complete intersection and then $U$ would be a local complete intersection in $Y$. However, that does not have to be the case.

Let's say that $N=3$, $d=1$, $k=1$. Or even more specifically, let $Y$ be a quadric cone, $p\in Y$ the vertex and $L\simeq \mathbb C\subset Y$ a line through $p$. Then $L$ is not a Cartier divisor on $Y$, so it cannot be "cut out" by a single equation. Actually this example may not be the best as $2L$ is a Cartier divisor.

So, let's take $N=4$, $d=2$, $k=1$, $Y$ the cone over $\mathbb P^1\times \mathbb P^1$, and $L\subset Y$ the cone over one of the rulings of the $\mathbb P^1\times \mathbb P^1$, then $L\simeq C^2$ and no multiple of $L$ is a Cartier divisor.

Actually I don't even see how you get a single disk to do what you claim at the vertex, but it is probably simply due to my limitations.

I don't think this can be done.

It seems to me that your $\widetilde U$ would be a submanifold of $\mathbb C^N$, so it should be a local complete intersection and then $U$ would be a local complete intersection in $Y$. However, that does not have to be the case.

Let's say that $N=3$, $d=1$, $k=1$. Or even more specifically, let $Y$ be a quadric cone, $p\in Y$ the vertex and $L\simeq \mathbb C\subset Y$ a line through $p$. Then $L$ is not a Cartier divisor on $Y$, so it cannot be "cut out" by a single equation. Actually this example may not be the best as $2L$ is a Cartier divisor.

So, let's take $N=4$, $d=2$, $k=1$, $Y$ the cone over $\mathbb P^1\times \mathbb P^1$, and $L\subset Y$ the cone over one of the rulings of the $\mathbb P^1\times \mathbb P^1$, then $L\simeq C^2$ and no multiple of $L$ is a Cartier divisor.