Let $K=\bar{K}$ a field, consider the line bundle $\mathcal O_{\mathbb P^2_K}(2)$ and global sections $s_0=x_0^2,s_1=x_1^2,s_2=x_0x_1,s_3=x_0x_2,s_4=x_2^2$ and $\phi:\mathbb P^2_K\to \mathbb P^4_K$ defined by $s_0,s_1,s_2,s_3,s_4$.
This is well defined on $\mathbb P^2_K$ and I want to determine if this is a closed immersion. To do so we can use II.7.2 in Hartshorne which does all the work. We have to show that writing $X=\mathbb P^2_K$, $X_{s_i}$ is affine $\forall i\in [0,4]$ and the morphisms $K[\frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\dots,\frac{x_{i+1}}{x_i},\dots,x_\frac{x_4}{x_i}]\to \Gamma(X_{s_i},\mathcal O_{X_{s_i}}),\quad \frac{x_j}{x_i}\mapsto \frac{s_j}{s_i}$ are surjective.
I showed that $X_{s_i}$ are affine. We have that $X_{s_0}=U_0\subseteq \mathbb P^2_K$, so I think we get $(U_0,\mathcal {O_{\mathbb P^2_K}}_{|X_{s_i}})\cong (\mathbb A^2,\mathcal O_{\mathbb A^2})$ and so we would get : $f:K[\frac{x_1}{x_0},\dots,\frac{x_4}{x_0}]\to K[x,y],\quad \frac{x_j}{x_0}\mapsto \frac{s_j}{s_0}$
My question is what is $\frac{s_j}{s_0}$ ? We would get something like $\frac{s_1}{s_0}=\frac{x_1^2}{x_0^2}$, but which variable corresponds to $x$ or $y$, what is a fraction like that doing in $K[x,y]$ ? Am I wrong with my previous identification ? With all these questions in mind I can't show that any such map is indeed surjective since I don't know if this map makes any sense.
This is typically not a closed immersion: at least in the characteristic not two case, $[0:1:1]$ and $[0:1:-1]$ both map to $[0:1:0:0:1]$. You'll need to spend a bit more time in the characteristic two case.
Let's discuss your question about $k[x,y]$. To check whether we have a closed immersion on $D(s_0)\subset\Bbb P^4$, we want to be looking at the ring map $$k\left[\frac{s_1}{s_0},\cdots,\frac{s_4}{s_0}\right] \to k\left[\frac{x_0^2}{x_0^2},\frac{x_0x_1}{x_0^2},\frac{x_0x_2}{x_0^2},\frac{x_1^2}{x_0^2},\frac{x_1x_2}{x_0^2},\frac{x_2^2}{x_0^2}\right] = k\left[\frac{x_1}{x_0},\frac{x_2}{x_0}\right],$$ and the identification with your $k[x,y]$ is typically $\frac{x_1}{x_0} \leftrightarrow x$ and $\frac{x_2}{x_0}\leftrightarrow y$. See section II.5 near proposition 11 for the relevant definitions if you're working out of Hartshorne - the key there is that for an $S$-module $M$ on $\operatorname{Proj} S$ and a homogeneous element $f$ of positive degree, we have $\widetilde{M}(D(f)) = M_{(f)}$, the homogeneous localization of $M$ at $f$. In our case, $M=S=k[x_0,x_1,x_2]$, $f=x_0^2$, so we can apply the definition to get the middle ring and then note that $\frac{x_0x_1}{x_0^2}=\frac{x_1}{x_0}$ and $\frac{x_0x_2}{x_0^2}=\frac{x_2}{x_0}$ are enough to generate the ring, giving the equality there.
It's not hard to see that this is indeed a surjective map of rings, but things will get trickier for other $i$.