Compute $I(U), I(V),I(U\cap V)$ and $I(U)+I(V)$. Let $U=Z(x^2-y)$ and $V=Z(y)$ in $k[x,y]$ where $k$ is a field. I recall that $$Z(I)=\{(x,y)\in k^2\mid \forall f\in I, f(x,y)=0\}$$
and $$I(U)=\{f\in k[x,y]\mid \forall (x,y)\in U, f(x,y)=0\}.$$

For $I(V)$, I have that $$f\in I(V)\iff f(x,0)=0\iff f\in (y).$$
We have that $I(U)=(x^2-y)$ but I don't understand why. Indeed,
$$(x,y)\in Z(x^2-y)\iff \forall f\in (x^2-y), f(x,y)=0\iff x^2-y=0\iff y=x^2,$$
therefore, $$Z(x^2-y)=\{(x,x^2)\mid x\in k\},$$
and thus $$f\in I(U)\iff f(x,x^2)=0$$
but how can I continue ?

For the rest, I think it's fine.
If $f(x,x^2)=0$ then $y=x^2$ is a root of $g(y)=f(x,y)\in F[x][y]$, therefore, $g(y)=(y-x^2)k(y)\in F[x][y]$ and thus $f(x,y)\in (y-x^2)$.
I wish to help you. we have theorem: $k[U]\cong k[x,y]/I(U) $, we have $k[U]=k[x,x^2]=k[x]$, then $k[x]\cong k[x,y]/I(U)\Rightarrow I(U)=(y-x^2).$ In similar way we have that $I(V)=(y)$.  and $I(U\cap V)=\sqrt{I(U)+I(V)}$. 
In your question $I(U\cap V)=I((0,0))=(x,y)$, and $I(U)+I(V)=(x^2-y,y)=(x^2,y)\Rightarrow \sqrt{I(U)+I(V)}=(x,y)$. we note that $I(U\cap V)\ne I(U)+I(V)$.