How would I correct this Hess cycle? Given
$$\ce{2 SO2(g) + O2(g) &lt;=&gt; 2 SO3(g)} \quad \Delta_\mathrm{r}H = \pu{-196.0 kJ mol^-1}$$
and $\Delta_\mathrm{f}H(\ce{SO2}) = \pu{-296.0 kJ mol^-1},$ find $\Delta_\mathrm{f}H(\ce{SO3}).$

I have chosen to solve this by constructing a Hess cycle and have come up with the following:

This method has lead me to the correct procedure of solving the equation: $$\Delta_\mathrm{r}H = -2\Delta_\mathrm{f}H(\ce{SO2}) + 2\Delta_\mathrm{f}H(\ce{SO3})$$
However, it is not entirely complete in my opinion and I am unsure on how to proceed in completing it, specifically on what I should place below the two arrows.
I am aware that it should be some combination of $\ce{S + O2}$ but am unaware on how exactly the coefficients would work.
You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps .
We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our convinience.
In your example, we disassociated $\ce{SO_2}$ molecule to corresponding standard state constituents and the enthalpy change associated with it will be $-\Delta H_f(\ce{SO_2})$.
$$\ce{2SO_2(g) + O_2(g) -&gt;[-2\Delta H_f(SO_2)] 2S(s) + 2O_2(g) + O_2(g)-&gt;[2\Delta H_f(SO_3)] 2SO_3(g)}$$
I took the following reaction through above mentioned 2 steps $$\ce{2SO_2(g) + O_2(g)-&gt;[\Delta_r H] 2SO_3(g)}$$
According to Hess law, the Enthalpy of reaction $\Delta_r H$ will be the summation of the all enthalpy changes occuring in the hypothetical intermediate steps.
$$\Delta_r H = -2\Delta H_f(\ce{SO_2}) + 2\Delta H_f(\ce{SO_3})$$