Let $X$ be a Banach space, $Y \subset X$ a subspace. Then $Y^{\bot \bot} = \overline{Y}^{w^*}$, where $A^\bot$ is the annihilator of $A$.
This confuses me. $Y^{\bot \bot}$ is a subspace of the bidual $X^{**}$, $w^*$-topology is a topology on the dual $X^*$ and $Y$ is a subset of $X$. How can I take a $w^*$-closure of a set from $X$, not from $X^*$? How can it be equal to a subset of the bidual?
You get the $w^*$-topology on $X^{**}$ by viewing $X^{**}$ as the dual space of the Banach space $X^*.$
Regarding your other questions, recall that there is a canonical embedding $J:X\hookrightarrow X^{**}$ that sends an $x\in X$ to the evaluation map at $x.$ More specifically, $J(x)=\operatorname{ev}_x$ where $\operatorname{ev}_x(\ell)=\ell(x)$ for $\ell \in X^*.$ In this way, it is common to identify a subset $Y\subset X$ with its image $J(Y)\subset X^{**}$ and this is often suppressed in the notation. If you want to explicitly include the embedding, you can write your statement as $Y^{\bot \bot} = \overline{J(Y)}^{w^*}.$