What you are missing is the concept of a sufficient statistic:
http://en.wikipedia.org/wiki/Sufficient_statistic
You have $P(F=f | \theta) = a(f|\theta)$, and $P(G=g|F=f)=b(g|f)$ the last one not depending on $\theta$. The distribution of your data $(F,G)$ is given by \begin{equation} P(F=f, G=g|\theta)=a(f|\theta) \cdot b(g|f) \end{equation} and the factorization theorem (see wikipedia article above) says directly that $F$ is a sufficient statistic for $\theta$, based on the sample $(F,G)$. Alternatively, we can use the definition directly: and calculate: \begin{equation} P(F=f,G=g| F=f,\theta)=\frac{P(F=f,G=g|\theta)}{P(F=f|\theta)}=b(g|f) \end{equation} which indeed do not depend on $\theta$. So, you do not need to consider the likelihood based on the full sample $(F,G)$, since you can always do as well using only the sufficient statistic.
An alternative answer is to point out that likelihood functions are not densities, they are only defined up to a multiplicative constant. So any two likelihood functions which have a ratio being independent of $\theta$, are equivalent in the sense that they will lead to equivalent inferences, when inference is based on the likelihood function. (The ratio in question in our case is of course $b(g|f)$.)