Question: How do you know the value of $\mathcal{F}_1$ and $\mathcal{F}_2$ in the proof below on (2) and (9)?
Let $Y = X_1 + X_2 + X_3$ where X_i is the outcome of the i-th toss of a fair coin. $X_i$ is IID with probability mass fucntion:
$$p_{X_i}(x) = \begin{cases}0.5 & x=1 \\ 0.5 & x=2 \end{cases}$$
And mean:
$$E[X_i] = 0.5$$
Verify the Martingale tower property, that is:
$$E\Big[~ E[Y~|~\mathcal{F}_n]~\Big|~\mathcal{F}_m\Big] = E[Y|F_m]$$
So I start the proof:
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = ?\tag{1}$$
$$\begin{matrix}F_2 = X_2, X_1&&F_1=X_1\end{matrix}\tag{2}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = E\Big[~ E[Y~|~X_2,X_1~\Big|~X_1\Big]\tag{3}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = E\Big[~ E[X_1+X_2+X_3~|~X_2,X_1~\Big|~X_1\Big]\tag{4}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = E\Big[~ X_1+X_2+E[X_3]~\Big|~X_1\Big]\tag{5}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = X_1 + E[X_2]+E[X_3]\tag{6}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = X_1 + 0.5+ 0.5\tag{7}$$
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = X_1 + 1\tag{7}$$
$$E[Y | F_1] = ?\tag{8}$$
$$F_1 = X_1\tag{9}$$
$$E[Y | F_1] = E[Y | X_1]\tag{10}$$
$$E[Y | F_1] = E[X_1 + X_2 + X_3 | X_1]\tag{11}$$
$$E[Y | F_1] = X_1 + E[X_2 + X_3]\tag{12}$$
$$E[Y | F_1] = X_1 + E[X_2] + E[X_3]\tag{12}$$
$$E[Y | F_1] = X_1 + 0.5 + 0.5\tag{13}$$
$$E[Y | F_1] = X_1 + 1\tag{14}$$
equating (7) and (14):
$$E\Big[~ E[Y~|~\mathcal{F}_2]~\Big|~\mathcal{F}_1\Big] = E[Y|F_1]$$
$\mathcal{F}_1$ and $\mathcal{F}_2$ are not functions, they are $\sigma$-algebras, i.e., sets of events. So, they don't assume values. I'll try to give you some intuition, but you need to go to a book and read about.
The idea behind $\mathcal{F}_1$ and $\mathcal{F}_2$ is that $\mathcal{F}_n$ has all the events you can measure until time $n$, so, it records events like $\{X_{n-1} \in A\}$ or $\{X_{n}, X_{n+1} \in B\}$, but it might not have information about $X_{n+1}$ or $X_{n+6}$, because those r.v. are on the future.
That way, $\mathcal{F}_n \subset \mathcal{F}_{n+1}$ since one $\mathcal{F}_{n+1}$ knows about time $n+1$ and about the past, we call this a filtration. A discrete time martingale is series of random variables $X_n$ such that $X_n$ is $\mathcal{F}_n$-measurable, that is, all the events we can read from $X_n$ (like $X_n \in [0,1]$, for example) belong to $\mathcal{F}_n$, and such that $\mathbb{E}(X_{n+1}| \mathcal{F}_n) = X_n$, i.e., your best prediction for $X_{n+1}$, knowing what you know today, is the value today.