2 Fixed one error

For $d\geq 5$, for a smooth plane curve of degree $d$, it is simply not true that the natural map $R_d \to H^1(X,T_X)$ is surjective: there are plenty of abstract deformations which are not plane curves. I am including below the proof that the natural map is surjective when $n\geq 4$, when $n=3$ and $d\leq 3$, and when $n=2$ and $d\leq 4$.

The basic idea is to look at the short exact sequence of sheaves $$0 \to T_X \to T_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(d)|_X \to 0.$$ Using the long exact sequence of cohomology, once you prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, then it follows that the connecting map $$H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X) \to H^1(X,T_X)$$ is surjective. Of course the image of $H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X)$ is $R_d$. Finally, using the long exact sequence of cohomology associated to the Euler sequence, $$0 \to \mathcal{O}_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus n+1}|_X \to T_{\mathbb{P}^n}|_X \to 0,$$ to prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, it suffices to prove that $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ and $h^2(X,\mathcal{O}_X)$ both equal $0$. These cohomologies can be computed from the usual computation of $H^q(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(r))$ using the long exact sequence of cohomology associated to the short exact sequence, $$0 \to \mathcal{O}_{\mathbb{P}^n}(r-d) \to \mathcal{O}_{\mathbb{P}^n}(r) \to \mathcal{O}_{\mathbb{P}^n}(r)|_X \to 0$$ For $n\geq 3$, always $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ equals $0$, and the second vanishing holds so long as $\text{dim}(X) \geq 3$, i.e., $n\geq 4$. It is not hard to prove that the second vanishing also holds if $n=3$ and $d\leq 3$. It definitely fails if $n=3$ and $d=4$, i.e., it fails for quartic K3 surface (although, in fact, the natural map is surjective again for $d\geq 5$).

This brings us to the case that $n=2$, i.e., $X$ is a plane curve. By adjunction and Serre duality on $X$, $h^1(X,\mathcal{O}_X(d))$ equals $0$. So the natural map is surjective if and only if $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$. By the Euler sequence again, this holds if and only if the map $$H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal{O}_X(1))^{\oplus 3}$$ is surjective. If $d$ equals $1$, $2$ or $3$, the second group is already zero. Finally, if $d$ equals $4$, your case, then by Serre duality the transpose map is $$H^0(X,\mathcal{O}_X)^{\oplus 3} \to H^0(X,\mathcal{O}_X(1))$$ So you are reduced to the question of whether the restriction map '$$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to H^0(X,\mathcal{O}_X(1))$$' is surjective, i.e., whether the next term $H^1(\mathbb{P}^2,\mathcal{O}(-3))$ equals $0$, which it does.

Once again, for $d\geq 5$, the natural map is not surjective.

Regarding your original question, how to "transform" a cocycle into an element of $R_d$: on the basic open affine covers of $\mathbb{P}^2$, write the corresponding Cech 1-cocycle for $T_{\mathbb{P}^2}|_X$ as the coboundary of Cech 0-chain0-cochain, which you can do since the cohomology group vanishes, and then . Then take the associated 0-chain image 0-cochain in $\mathcal{O}(d)|_X$. This The coboundary of this 0-cochain will be the image of your original 0-cochain in $\mathcal{O}_X(d)$, which is zero since the composite map $T_X \to \mathcal{O}_X(d)$ is zero. So your 0-cochain is a 0-cocycle, i.e., a global section in $R_d$.

For $d\geq 5$, for a smooth plane curve of degree $d$, it is simply not true that the natural map $R_d \to H^1(X,T_X)$ is surjective: there are plenty of abstract deformations which are not plane curves. I am including below the proof that the natural map is surjective when $n\geq 4$, when $n=3$ and $d\leq 3$, and when $n=2$ and $d\leq 4$.
The basic idea is to look at the short exact sequence of sheaves $$0 \to T_X \to T_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(d)|_X \to 0.$$ Using the long exact sequence of cohomology, once you prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, then it follows that the connecting map $$H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X) \to H^1(X,T_X)$$ is surjective. Of course $H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X)$ is $R_d$. Finally, using the long exact sequence of cohomology associated to the Euler sequence, $$0 \to \mathcal{O}_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus n+1}|_X \to T_{\mathbb{P}^n}|_X \to 0,$$ to prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, it suffices to prove that $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ and $h^2(X,\mathcal{O}_X)$ both equal $0$. These cohomologies can be computed from the usual computation of $H^q(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(r))$ using the long exact sequence of cohomology associated to the short exact sequence, $$0 \to \mathcal{O}_{\mathbb{P}^n}(r-d) \to \mathcal{O}_{\mathbb{P}^n}(r) \to \mathcal{O}_{\mathbb{P}^n}(r)|_X \to 0$$ For $n\geq 3$, always $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ equals $0$, and the second vanishing holds so long as $\text{dim}(X) \geq 3$, i.e., $n\geq 4$. It is not hard to prove that the second vanishing also holds if $n=3$ and $d\leq 3$. It definitely fails if $n=3$ and $d=4$, i.e., it fails for quartic K3 surface (although, in fact, the natural map is surjective again for $d\geq 5$).
This brings us to the case that $n=2$, i.e., $X$ is a plane curve. By adjunction and Serre duality on $X$, $h^1(X,\mathcal{O}_X(d))$ equals $0$. So the natural map is surjective if and only if $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$. By the Euler sequence again, this holds if and only if the map $$H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal{O}_X(1))^{\oplus 3}$$ is surjective. If $d$ equals $1$, $2$ or $3$, the second group is already zero. Finally, if $d$ equals $4$, your case, then by Serre duality the transpose map is $$H^0(X,\mathcal{O}_X)^{\oplus 3} \to H^0(X,\mathcal{O}_X(1))$$ So you are reduced to the question of whether the restriction map '$$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to H^0(X,\mathcal{O}_X(1))$$' is surjective, i.e., whether the next term $H^1(\mathbb{P}^2,\mathcal{O}(-3))$ equals $0$, which it does.
Once again, for $d\geq 5$, the natural map is not surjective.
Regarding your original question, how to "transform" a cocycle into an element of $R_d$: on the basic open affine covers of $\mathbb{P}^2$, write the corresponding Cech 1-cocycle for $T_{\mathbb{P}^2}|_X$ as the coboundary of Cech 0-chain, which you can do since the cohomology group vanishes, and then take the associated 0-chain in $\mathcal{O}(d)|_X$. This will be a 0-cocycle, i.e., a global section in $R_d$.