$R$ and $Q$ are two stopping times for a filtration $(\mathcal{F}_k)_k$ and $ X \in L^1.$ We can prove that (and will be used below in the proof): $E[E[X|\mathcal{F}_{Q}]|\mathcal{F}_R]=E[X|\mathcal{F}_{\min(R,Q)}]=E[E[X|\mathcal{F}_R]|\mathcal{F}_Q]$
We want to prove that the following holds, almost surely, $$E[X|\mathcal{F}_{\max(R,Q)}]+E[X|\mathcal{F}_{\min(R,Q)}]=E[X|\mathcal{F}_R]+E[X|\mathcal{F}_Q].$$
One way to prove it:
$\mathcal{F}_{\max(R,Q)}=\sigma(\mathcal{F}_R \cup \mathcal{F}_Q)=\sigma(\{K_1 \cap K_2, (K_1,K_2) \in \mathcal{F}_Q \times \mathcal{F}_R\}).$
$E[X|\mathcal{F}_R],E[X|\mathcal{F}_Q],E[X|\mathcal{F}_{\max(R,Q)}],E[X|\mathcal{F}_{\min(R,Q)}]$ are all $\mathcal{F}_{\max(R,Q)}$-measurable.
So it's sufficient to prove that for $K_1 \in \mathcal{F}_{R},K_2 \in \mathcal{F}_{Q},$ $$E[X1_{K_1 \cap K_2 }]+E[1_{K_1 \cap K_2}E[X|{\mathcal{F}}_{\min(R,Q)}]]=E[1_{K_1 \cap K_2}E[X|\mathcal{F}_R]]+E[1_{K_1 \cap K_2}E[X|\mathcal{F}_Q]] \ \ \ \ \ \ \ \ (1)$$
We have :
$E[X1_{K_1 \cap K_2 }]+E[1_{K_1 \cap K_2}E[X|{\mathcal{F}}_{\min(R,Q)}]]=E[X1_{K_1 \cap K_2 \cap \{R \leq Q\} }]+E[X1_{K_1 \cap K_2 \cap \{R > Q\} }]+E[1_{K_1 \cap K_2 \cap {\{R \leq Q\}}}E[X|{\mathcal{F}}_{\min(R,Q)}]]+E[1_{K_1 \cap K_2\{R>Q\}}E[X|\mathcal{F}_{\min(R,Q)}]].$
Since $K_1 \cap K_2 \cap \{R \leq Q\} \in \mathcal{F}_Q,K_1 \cap K_2 \cap\{R>Q\} \in \mathcal{F}_R,$ then
$E[X1_{K_1 \cap K_2 \cap \{R \leq Q\} }]=E[1_{K_1 \cap K_2 \cap \{R \leq Q\} }E[X|\mathcal{F}_Q]],$
$ E[X1_{K_1 \cap K_2 \cap \{R > Q\} }]=E[1_{K_1 \cap K_2 \cap \{R > Q\} }E[X|\mathcal{F}_R]],$
$ E[1_{K_1 \cap K_2 \cap {\{R \leq Q\}}}E[X|{\mathcal{F}}_{\min(R,Q)}]]=E[1_{K_1 \cap K_2 \cap {\{R \leq Q\}}}E[X|{\mathcal{F}}_{R}]]$,
$ E[1_{K_1 \cap K_2\{R>Q\}}E[X|\mathcal{F}_{\min(R,Q)}]]=E[1_{K_1 \cap K_2\{R>Q\}}E[X|\mathcal{F}_{Q}]]$
In other word $(1)$ holds, to conclude using $\pi$-$\lambda$ theorem.
Since the above is true, the following should also hold (by induction...): $R_1,...,R_d$ are stopping times for a filtration $(\mathcal{F}_n)_n$ then $$E[X|\mathcal{F}_{\max_{1 \leq k \leq d}R_k}]=\sum_{k=1}^d(-1)^{k-1}\sum_{1 \leq i_1<...<i_k \leq d}E[X|\mathcal{F}_{\min_{ 1 \leq p \leq k} R_{i_p}}]$$