This is related to Iitaka, Algebraic Geometry, Chpt 1, sec 18 b).
Let $W_i=D(T_i)$ of $\mathrm{Spec}(R[T_0,\dots, T_n])$ for all $i$. Then there is inclusion map $R[T_0/T_i,\dots ,T_n/T_i]\to R[T_0,\dots, T_n,1/T_i]$. Note that the former ring is open set of $T_i\neq 0$ of $P^n_R$. So this induces $W_i\to P^n_R$ with image lying in $T_i\neq 0$ of $P^n_R$.
$\textbf{Q:}$ If $R$ is a algebraically closed field, then I would expect $\mathrm{Spec}(R[T_0,\dots, T_n,1/T_i])\to \mathrm{Spec}(R[T_0/T_i,\dots, T_n/T_i])$ is epimorphism as topological space where I will forget about sheaves. Is there an easy way to see this is topological homeomorphism? It is not clear whether I can even apply integrality argument here.
For simplicity let $i = 0$. Notice that $R[T_0/T_0,\dots ,T_n/T_0]= R[T_1/T_0,\dots ,T_n/T_0]:=S$ and $R[T_0,\dots, T_n,1/T_0] = S[T_0,\frac1{T_0}]$. Under $S \to S[T_0,\frac1{T_0}]$, the preimages of the primes of the form $T_0 - a$, where $a \in R \setminus \{0\}$ are all $(0)$. Or one can compare the Krull dimension of these two rings.
Thus, the map on the spectra is not a homeomorphism.