We can to understand $\Bbb{P}(A \vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$.
In that sense one would interpret $\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega)$ for $\omega \in B \cap C$
$$\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[\Bbb{E}[1_A |\sigma(1_B, 1_C)] |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B, 1_C)] \vert \sigma(1_C)](\omega) = \Bbb{E}[1_A |\sigma(1_B, 1_C)](\omega) = P(A \vert B,C) $$
Note $\Bbb{E}[1_A\vert \sigma(1_B)]$ is a random variable
$$\Bbb{E}[1_A\vert \sigma(1_B)] = P(A\vert B) 1_B + P(A\vert B^c)1_{B^c} $$
The relation we used for conditional expectations was this one.
$$\mathcal{G} \subset \mathcal{F} \Rightarrow\Bbb{E}[\Bbb{E}[1_A |\mathcal{G}] |\mathcal{F}] = \Bbb{E}[\Bbb{E}[1_A |\mathcal{G}]$$
where $\mathcal{F},\mathcal{G}$ are sigma algebras.
Note that we are understanding $P(A\vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)]$ wich is different than $\Bbb{E}[1_A |\sigma(1_B 1_C)]$ to finally get a number we write
$$\Bbb{E}[1_A |\sigma(1_B, 1_C)] = P(A\vert B \cap C)1_{B \cap C} + P(A\vert B^c \cap C)1_{B^c \cap C} + P(A\vert B \cap C^c)1_{B \cap C^c}+ P(A\vert B^c \cap C^c)1_{B^c \cap C^c}$$
So you define $P(A \vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)] (\omega)$ for any $\omega \in B \cap C$
As you would have defined $P(A\vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$