Tossing a coin once

Let $S_{1}=\{h, t\}$ be the sample space or the set of elementary events, which are $h$ for head and $t$ for tail.
The probability distribution is
$p:p_{1}(S_{1})\rightarrow [0, +\infty)$ ($p_{1}(S_{1})$ power set of $S_{1}$) and we have that

$p_{1}(\emptyset)=0$
$p_{1}(\{h\})=p_{1}(\{t\})=1/2$
$p_{1}(S_{1})=1$

Tossing a coin twice

$S_{2}=\{hh, ht, th, tt\}$.
$p_{2}(\emptyset)=0$
$p_{2}(\{hh\})=p_{2}(\{ht\})=p_{2}(\{th\})=p_{2}(\{tt\})=1/4$.
$p_{2}(S_{2})=1$

Now, this is easy but I don't understand independent events.
https://en.wikipedia.org/wiki/Probability_space#Independence

Looking at the probability space of a single tossing we have $4\cdot 4=16$ couples we can place in the formula $p_{1}(\cdot , \cdot )=p_{1}(\cdot)p_1({\cdot})$ and by checking all of them I have that

$A, B$
$\emptyset, \emptyset$
$\emptyset, h$
$\emptyset, t$
$\emptyset, S$

$S, \emptyset$
$S, h$
$S, t$
$S, S$

satisfy the condition.

1. What's the point of this?
2. How does the double coin tossing fit into it.
3. How do I actually calculate probability in the double coin tossing given that it is already given with the function $p_{2}$.

Intuitively I understand that $p_{2}(hh)=1/4=1/2\cdot 1/2=p_{1}(h)p_{1}(h)$ but again I don't get how this fits with the two probability spaces.
This latter check is on different probability spaces.