A proper wanna-be generalization of the "tower property" of conditional expectations would be:
$$\mathbb E(\mathbb E(X\mid \mathcal F) \mid \mathcal G) = \mathbb E(X \mid \mathcal F\cap \mathcal G)$$
regardless of, whether the $\sigma$-algebras $\mathcal F$ and $\mathcal G$ can be compared with $\subseteq$.
My guess is this is false. What is a counterexample? (or proof?)
Consider a sample space with three outcomes $\Omega = \{p,q,r\}$, each having probability $1/3$. Let $A = \{p,q\}$ and $\mathcal{F} = \sigma(\{A\}) = \{\emptyset, \Omega, \{p,q\}, \{r\}\}$. Let $B = \{q,r\}$ and $\mathcal{G} = \sigma(\{B\}) = \{\emptyset, \Omega, \{q,r\}, \{p\}\}$. Note that $\mathcal{F} \cap \mathcal{G} = \{\emptyset, \Omega\}$ is the trivial $\sigma$-field.
Let $X = 1_A$. Then $E[X \mid \mathcal{F}] = X$, and we have $$E[E[X \mid \mathcal{F}] \mid \mathcal{G}]] = E[X \mid \mathcal{G}] = P(A \mid B) 1_B + P(A \mid B^c) 1_{B^c}.$$ Now $P(A \mid B) = 1/2$ and $P(A \mid B^c) = 1$ so this is not constant. But since $\mathcal{F} \cap \mathcal{G} = \{\emptyset, \Omega\}$, we have $E[X \mid \mathcal{F} \cap \mathcal{G}] = E[X] = 2/3$.