From Working with Weighted Complete Intersections by Iano-Fletcher, which appears to be available freely here:
Let $f = x^5 + y^3 + z^2 \in k[x,y,z]$ with weights $6,10$, and $15$ respectively. Define $X: (f = 0 ) \subseteq \mathbb{P} = \mathbb{P}(6,10,15)$. (...) $$\mathbb{P}(6,10,15) \simeq \mathbb{P}(6,2,3) \simeq \mathbb{P}(3,1,3) \simeq \mathbb{P}(1,1,1)$$ The monomials transform as $$(x^5, y^3, z^2) \mapsto (x, y^3, z^2) \mapsto (x, y^3, z) \mapsto (x,y,z)$$ Thus $X \subset \mathbb{P} \simeq \{x + y + z = 0\} \subseteq \mathbb{P}^2 = \mathbb{P}^1 \subseteq \mathbb{P}^2$. Of course the coordinate rings of the affine cones over $X \subseteq \mathbb{P}$ and $\mathbb{P}^1 \subseteq \mathbb{P}^2$ are not isomorphic.
First, note that this is exactly your case. Multiplying all of the weights by two amounts to making the $\mathbb{G}_m$ action non-effective (the subgroup $\{\pm 1\} \subseteq \mathbb{G}_m$ stabilizes everything).
I think what you missed above (and which confused me thinking about it as well) is that "$x$", "$y$", etc. don't mean the same thing after each isomorphism. Let's do the first one explicitly, and the others will be similar.
We know that $\mathbb{P}(6,10, 15) = \text{Proj}(k[x,y,z])$ with the grading $|x| = 6, |y| = 10, |z| = 15$ is isomorphic to $\mathbb{P}(6, 2, 3)$. However, this isomorphism is not as simple as the isomorphism of graded rings showing $\mathbb{P}(6,10,15) \simeq \mathbb{P}(12, 20, 30)$ (for example). It's somewhat non-canonical (this has something to do with the fact that the affine cones of the resulting varieties aren't isomorphic). Let $S$ be the first graded ring and $T$ the second. Now, we know that $\text{Proj}(S) \simeq \text{Proj}(S^{(5)})$. How does this isomorphism act on closed sets? (This is all spelled out in EGA II.2.4.7). The isomorphism is induced by the inclusion $S^{(5)} \hookrightarrow S$. Under this map, the homogeneous primes of $S$ containing an element $f$ of degree $d$ correspond exactly to the homogeneous primes of $S^{(5)}$ containing $f^5$ (note $V_+(f) = V_+(f^d)$!). Our element, $x^2 + y^3 + z^5$ is actually homogeneous of degree $30$, so $x^2 + y^3 + z^5 \in S^{(5)}$ and thus, in $S^{(5)}$, our homogeneous ideal is just the ideal of $S^{(5)}$ generated by the same polynomial.
$S^{(5)}$ is generated by monomials $x^a y^b z^c$ with $15a + 10b + 6c = 5n$ for some $m$. Since $5$ doesn't divide $6$, this means that $5|c$, and thus any such monomial is of the form $x^{a} y^b z^{5c'}$. Since any monomial of this form has degree a multiple of $5$, we conclude that these span $S^{(5)}$. This shows that $S^{(5)}$ is $k[x, y, z^5]$ where $x$ has weight $15$, $y$ has weight $10$, and $z^5$ has weight $30$. This is isomorphic to the graded ring $k[s, t, r]$, up to an overall multiplicative factor in the grading, where $s$ has weight $3$, $t$ has weight $2$, and $r$ has weight $6$, by the map $s \mapsto x, t \mapsto y, r \mapsto z^5$. This map induces a canonical isomorphism on their $\text{Proj}$ (which, if you recall, is constructed by considering degree $0$ fractions, so multiplying the degree of everything by a constant doesn't change this). In particular, the isomorphism $\text{Proj}(S^{(5)}) \simeq \text{Proj}(T)$ takes $V_+(x^2 + y^3 + z^5)$ to $V_+(s^2 + t^3 + r)$. Note that this new polynomial, $s^2 + t^3 + r$ is homogeneous of degree $6$ in $T$. Continuing this way, we finally get an isomorphism of $\text{Proj}(S)$ with $\text{Proj}(k[x,y,z])$ where the latter has the usual $(1,1,1)$ grading. This isomorphism takes $x^2 + y^3 + z^5$ to $x + y + z$, which is of course a homogenous linear function in three variables. Thus, $V_+(x^2 + y^3 + z^5) \subseteq \mathbb{P}(15,10,6)$ is isomorphic to $V_+(x+y+z) \simeq \mathbb{P^1} \subseteq \mathbb{P}(1,1,1) = \mathbb{P}^2$.
Notice that what I did here is essentially the same as the algebra you did above. However, there are a few important differences to point out:
First, it appears that some of the confusion stems from the fact that you used the letters $R$, $x,y,z$ to represent different but isomorphic things.
Second, it is not the case that $R \simeq R^{(d)}$ for any $d$. It is only the case that $\text{Proj}(R) \simeq \text{Proj}(R^{(d)})$. Note that the affine cones over our initial and final varieties are very different! The former is a singular hypersurface of degree $5$, while the latter is just a plane.
Finally, I'd like to go back to my earlier comment, which is interesting in light of the understanding that we're really just dealing with $\mathbb{P}^1$. First of all, you're completely correct in the interpretation that $\text{Proj}(k[x,y,z])/f(x,y,z)$ is just the quotient of $V(f(x,y,z)) \subset \mathbb{A}^3$ by the action of the circle $\mathbb{G}_m$. In this case, the action is $\zeta: (x,y,z) \mapsto (\zeta^{15} x, \zeta^{10}y, \zeta^{5}z)$. Indeed, it's true (and I don't know a good reference for this fact; I'd appreciate if someone finds one) that the Proj of any graded algebra is the quotient of its spectrum by a circle action.
This interpretation is very interesting topologically. Knowing that the quasi-homogeneous variety is just $\mathbb{P}^1$ tells us that the link of the singularity $L:= \{(x,y,z) \in S^5 \subseteq \mathbb{C}^3 \mid x^2 + y^3 + z^5 = 0\}$ admits a fixed point-free circle action such that the quotient is just $S^2$! This circle action is "almost" a fiber bundle; the reason it isn't is that at points when, for example, $z = 0$, the fiber $\{(\zeta^{15}x, \zeta^{10}y, 0)\}$ has multiplicity $5$. Topologists refer to such $3$-manifolds as Seifert fiber spaces, and they form interesting examples of $3$-manifolds whose algebraic topology is generally manageable. Indeed, knowing the genus of the quotient surface in addition to some information about orientability (which is trivial in this case since everything is complex) and the local description of the fibers is almost enough to determine the fundamental group and homology entirely. In fact, this particular $3$-manifold is very famous! It's the Poincaré Homology Sphere, the earliest historical example of a manifold with the homology groups of $S^3$ which is not homeomorphic to it.
Best Answer
There is a very ampleness condition due to Delorme in [Del75a]. You can also look at [BR86] for an English reference. We have changed the dimension $n$ in your notation to an $r$ to closer match the notation in these references.
We start with some definitions.
Definition [Del75a, Def. 2.1; BR86, Defs. 4B.1, 4B.2, and 4B.3]. We let $m_J := \operatorname{lcm}\{a_i \mid i \in J\}$ for every non-empty subset $J \subseteq \{0,1,\ldots,r\}$, and let $m := m_{\{1,2,\ldots,r\}}$. Now set $$G := \begin{cases} -a_r & \text{if $r = 0$, and}\\ \displaystyle-\sum_{i=0}^r a_i + \frac{1}{r}\sum_{2 \le \nu \le r+1} \binom{r-1}{\nu-2}^{-1} \sum_{\lvert J \rvert = \nu} m_J & \text{otherwise.} \end{cases}$$ We then say that an integer $n \ge 0$ satisfies condition $D(n)$ if one of the following equivalent conditions hold:
Given a relation $\sum_{i = 0}^r B_ia_i = n + km$ with $k \in \mathbf{Z}_{>0}$ and $B_i \in \mathbf{Z}_{\ge0}$ for every $i$, there exist $b_i \in \mathbf{Z}_{\ge0}$ with $B_i \ge b_i$ for every $i$, such that $\sum_{i = 0}^r b_ia_i=km$.
Every monomial $\prod_{i=0}^r x_i^{B_i}$ of degree $n+km$ is divisible by a monomial $\prod_{i=0}^r x_i^{b_i}$ of degree $km$.
We then define $F$ to be the smallest integer such that $n > F$ implies $D(n)$ holds. We also define $E$ to be the smallest integer such that $n > E$ implies $D(mn)$ holds. Note that $mE \le F$.
One can then show that $F$ is finite and $F \le G$ [Del75a, Prop. 2.2; BR86, Prop. 4B.5]. The proof is a double induction on $k$ and $r$, and boils down to the pigeon-hole principle.
With notation as above, we then have the following:
Theorem [Del75a, Prop. 2.3; BR86, Thm. 4B.7]. Let $X = \mathbf{P}(a_0,a_1,\ldots,a_r)$ be a weighted projective space over a commutative ring $A$, where $a_i \ge 1$ for every $1$. With notation as above, we have the following:
$\mathcal{O}_X(m)$ is an ample invertible sheaf.
If $n > F$, then the sheaf $\mathcal{O}_X(n)$ is globally generated.
If $n > 0$ and $n > E$, then the sheaf $\mathcal{O}_X(nm)$ is very ample.
We prove (3), since this is what you are interested in. For every $p \in \mathbf{Z}_{>0}$, the condition $D(mn)$ implies that every monomial of degree $pmn = (p-1)mn + mn$ in $A[x_0,x_1,\ldots,x_r]$ is divisible by a monomial of degree $(p-1)mn$. Thus, the $mn$th Veronese subring $A[x_0,x_1,\ldots,x_r]^{(mn)}$ of $A[x_0,x_1,\ldots,x_r]$ is generated in degree $1$ over $A$.
You can therefore embed $X$ into $\mathbf{P}^N_A$, where $N$ is the number of generators of $A[x_0,x_1,\ldots,x_r]_{mn}$, by $$X = \operatorname{Proj}\bigl(A[x_0,x_1,\ldots,x_r]\bigr) \simeq \operatorname{Proj}\bigl(A[x_0,x_1,\ldots,x_r]^{(mn)}\bigr) \hookrightarrow \mathbf{P}^N_A.$$ Unraveling the definitions given above, we note that in particular, setting $$n = \biggl\lfloor\frac{1}{m}G\biggr\rfloor+1$$ works.
References
[BR86] Mauro Beltrametti and Lorenzo Robbiano, "Introduction to the theory of weighted projective spaces," Exposition. Math. 4 (1986), no. 2, 111–162. mr: 879909.
[Del75a] Charles Delorme, "Espaces projectifs anisotropes," Bull. Soc. Math. France 103 (1975), no. 2, 203–223. doi: 10.24033/bsmf.1802. mr: 404277.
[Del75b] Charles Delorme, "Erratum: 'Espaces projectifs anisotropes'," Bull. Soc. Math. France 103 (1975), no. 4, 510. doi: 10.24033/bsmf.1812. mr: 404278.