You initially asked about radiation from the gas (i.e., emitted radiation, as would be relevant in the case of carbon dioxide or water vapor, for example, at very high temperatures) but then added an expression for radiative heat transfer from the outer wall through the gas. So I'll assume you meant the latter—that the gas itself is not radiating significantly but that the heat of the outer wall is so high that radiation from it is relevant.
You cannot add the heat flux in your first equation to the heat flux in your second equation because they were derived under different conditions. Specifically, the first equation was derived under the condition that only convective heat transfer affects the outside of the steel tube. Let's look at this derivation.
The heat flow from the steel tube to the water is
$$Q=2\pi r_1Lh_1(T_1-T_\mathrm{water}).$$
The heat flow through the inner tube is
$$Q=\frac{2\pi Lk_\mathrm{steel}(T_2-T_1)}{\ln(r_2/r_1)}.$$
The heat flow from the gas and wall to the steel tube, with the radiation mechanism added, is
$$Q=2\pi r_2 L\left(h_2(T_\mathrm{gas}-T_2)+\frac{\sigma(T_\mathrm{wall}^4-T_2^4)}{\frac{1}{\epsilon_\mathrm{steel}}+\frac{1-\epsilon_\mathrm{wall}}{\epsilon_\mathrm{wall}}\left(\frac{r_2}{r_\mathrm{wall}}\right)}\right).$$
(The radiation term should be equivalent to yours; it's derived, for example, in Incropera's table "Special Diffuse, Gray, Two-Surface Enclosures".)
These terms are all equal at steady state. Now, if the radiation term weren't present, then you could take advantage of the fact that these equivalent heat fluxes are linear in $T$ and construct an equivalent thermal resistance, as you do in your question.
As far as I know, however, the presence of the radiation term precludes a tidy analytical solution for the general case.
Nevertheless, if you can make two assumptions, you can recover the simple thermal resistance form. One is that $T_\mathrm{wall}\approx T_\mathrm{gas}$, and the other is that $T_\mathrm{wall}$ isn't much higher than $T_2$. In such a case, you can linearize $T_\mathrm{wall}^4-T_2^4$ as $4T_\mathrm{wall}^3(T_\mathrm{wall}-T_2)$, turning the third equation above into
$$Q=2\pi r_2 L\left(h_2+\frac{4\sigma T_\mathrm{wall}^3}{\frac{1}{\epsilon_\mathrm{steel}}+\frac{1-\epsilon_\mathrm{wall}}{\epsilon_\mathrm{wall}}\left(\frac{r_2}{r_\mathrm{wall}}\right)}\right)(T_\mathrm{wall}-T_2)=2\pi r_2 L\left(h_2+R\right)(T_\mathrm{wall}-T_2),$$
where $R$ captures all that radiative heat transfer information. In this case, the heat transfer from wall to water per unit length is
$$\frac{Q}{L}=\frac{2\pi r_1(T_\mathrm{wall}-T_\mathrm{water})}{\frac{1}{h_1}+\frac{r_1}{(h_2+R)r_2}+\frac{\ln(r_2/r_1)}{k}}.$$
