The sigma-algebra generated by $1_{[0,1/2]}$ is simply
$$
\bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}.
$$
It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomain of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)
The situation for $1_{[1/4,3/4]}$ is similar.
The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.
The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.
Check that both sides are equal!
Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?
If $f: X \to (Y,\Sigma)$ is a function from a set to any measurable space (a space equipped with a $\sigma$-algebra) then it is easy to check that $\{f^{-1}(S)\,:\,S \in \Sigma\} = \sigma(f)$ is a $\sigma$-algebra (sorry, I can't bring myself to writing $\{f \in \Sigma\}$ for the pre-image $f^{-1}(S)$ of $S$ even if there's some justification). This is because taking pre-images commutes with unions, intersections and complements. By definition of measurability, $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ on $X$ making $f$ measurable, i.e. $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ such that $f^{-1}(S) \in \mathcal{A}$ for all $S \in \Sigma$. So the answer to your first question is no.
Now if $f,g: X \to (Y,\Sigma)$ are two functions, then $\sigma(f,g)$ is the smallest $\sigma$-algebra making both $f$ and $g$ measurable. It doesn't have such an easy description in general, but it is clear that $\sigma(f), \sigma(g) \subset \sigma(f,g)$ and actually $\sigma(f,g)$ is the intersection of all $\sigma$-algebras containing both $\sigma(f)$ and $\sigma(g)$. In fact, if $\mathcal{S}$ is any collection of subsets of $X$ then it generates a $\sigma$-algebra $\sigma(\mathcal{S}) = \bigcap_{\mathcal{S} \subset \mathcal{A}} \mathcal{A}$, where the intersection is taken over all $\sigma$-algebras containing $\mathcal{S}$. Since the power set of $X$ contains $\mathcal{S}$ and is a $\sigma$-algebra, this intersection is non-empty. You should convince yourself that an arbitrary intersection of $\sigma$-algebras is again a $\sigma$-algebra.
In your example, you have $X = Y^{\mathbb{N}}$ and in this situation the $\sigma$-algebra $\sigma(p_{1},\ldots,p_{n})$ has a semi-concrete description. Note that a pre-image of $p_{i}$ is of the form $\underbrace{Y \times \cdots \times Y}_{(i-1)\;\text{times}} \times S_{i} \times Y \times Y \times \cdots$, where $S_{i} \subset Y$ is arbitrary. By taking the intersection of the sets $p_{i}^{-1}(S_{i})$ this means that all the sets of the form $S_{1} \times \cdots \times S_{n} \times Y \times Y \times \cdots$ with $S_{1},\ldots,S_{n} \in \Sigma$ must belong to $\sigma(p_{1},\cdots,p_{n})$ and in fact, these so-called cylinder sets generate the $\sigma$-algebra $\sigma(p_{1},\cdots,p_{n})$.
Best Answer
The definition of the sigma alebra generated by a random variable $Y$ on a finite universe $\Omega$ is $$ \sigma(Y) = \{ Y^{-1}(A) : A \in \mathcal P(Y(\Omega)) \} $$
so given the universe $\Omega = \{ HH, HT, TH, TT\}$ and the definition of $S_1$ and $S_2$ you clearly have
$$ \sigma(S_1) = \{\emptyset, \{ HH, HT\}, \{ TH, TT\} ,\Omega\} $$ $$ \sigma(X) = \{\emptyset,\{HT, TH \}, \{HH, TT \}, \Omega\} $$
because
$X(\Omega) = \{0,1\}, \quad \mathcal P(X(\Omega)) = \{\emptyset, \{0\},\{1\},\{0,1\} \}$
$X^{-1}(\{ 1 \}) = \{HT, TH\}, \quad X^{-1}(\{ 0 \}) = \{HH, TT\}$