Is weak forcing a semantic relation? I'm vaguely recalling things from innumerable aeons ago:
A condition $p$ forces the negation $\sim\varphi$ of a sentence $\varphi$ precisely if no condition that extends $p$ forces $\varphi$.  From this it can follow that $p\Vdash \sim\sim\exists x\ \varphi(x)$ even though $p\not\Vdash \exists x\ \varphi(x)$.  Things like this seem unpleasant at first.  The fact that both $p\Vdash \varphi$ and $p\not\Vdash \psi$ can hold even though the models of $\varphi$ are precisely the models of $\psi$ is what I would mean by saying forcing is not a semantic relation: it depends on things about $\varphi$ and $\psi$ other than the meanings of $\varphi$ and $\psi$.
If I recall correctly, "$p$ weakly forces $\varphi$" means $p$ forces ${\sim\sim}\varphi$.
Is weak forcing a semantic relation in the sense that if $\varphi$ and $\psi$ have the same models then a condition that forces one of them also forces the other?
 A: The forcing relation that Boolos, Burgess, and Jeffrey define is what might be called "strong forcing". This is a very concrete relation, which is easy to define in the ground model, but it does not in general respect classical logical equivalence. The logic that is preserved is an intuitionistic one (by "preserved" I mean the question of which inference rules are sound when provability is replaced by forcing). 
One thing that is somewhat confusing is that many set theory books now write $\Vdash$ for the semantic relation: $p \Vdash \phi$ means that for every generic $G$ containing $p$, $M[G]$ satisfies $\phi$. This relation clearly preserves classical logic, but on its face there is no way to see that it is definable in the ground model. 
To prove that, many contemporary books use some decorated notation (e.g. $\Vdash^*$ or $\Vdash_s$) to define an auxiliary relation (the one that Boolos, Burgess, and Jeffrey call $\Vdash$) which is obviously definable in the ground model but which does not obviously have the logical properties of the semantic forcing relation. 
In any case what has to be proved is that there is some forcing relation definable in the ground model that has the property:

If $G$ is a generic filter, then $M[G]$ satisfies a formula $\phi$ if and only if there is some condition in $G$ that forces $\phi$

In my version of BB&J this is Lemma 23.7. Their exposition is focused on one specific forcing notion, Cohen forcing, which allows them to mix the generic filter $G$ with the generic set, which they call $A$, that is obtained from the filter.  So it takes a minute to reconstruct the general results from the specialized statements of their lemmas. 
You are right that you can define a "weak" forcing relation $\Vdash_w$ from the strong one $\Vdash_s$ with the rule $p \Vdash_w \phi \Leftrightarrow p\Vdash_s \lnot\lnot \phi$, and then the weak one will preserve classical logic. 
The key point is that if $p \Vdash_s \lnot \lnot \phi$ then the set of $r \leq p$ that force $\phi$ is dense below $p$, and thus any generic filter that includes $p$ has to include such a $r$. Thus if an entire generic filter forces $\lnot \lnot \phi$ then that filter also forces $\phi$, even with the syntactic definition of forcing.  Similarly, it follows from the definition of $\Vdash_s$ that if $p \Vdash_s \lnot \lnot \lnot \lnot \phi$ then $p \Vdash_s \lnot \lnot \phi$, which should not be unexpected if we think of $\Vdash_s$ as having a sort of intuitionistic logic. 
In books like Kunen's, the $\Vdash^*$ relation is a little different than the $\Vdash_s$ relation as defined by BB&J. Kunen's definition appeals to dense sets in the clause for $(\exists x)$, while the "strong" definition requires that a condition that forces an existential quantifier has to force a particular witness at the same time. So you can view the $\Vdash^*$ type definition as directly incorporating a double negation, which is how Kunen is able to prove $p \Vdash \phi \Leftrightarrow p \Vdash^* \phi$. 

Part 2
Let $\Vdash_s$ be the strong forcing relation from BB&J. We have from BB&J:

Lemma 23.7. If $G$ is a generic filter and $\phi$ is a sentence then $M[G]$ satisfies $\phi$ if and only if there is a $p \in G$ with $p \Vdash_s \phi$. 

Now the solution to your question is essentially a little exercise using the key point I mentioned above. 

Proposition 1. $p \Vdash_s \lnot\lnot \phi$ if and only if for every generic $G$ containing $p$, $M[G]$ satisfies $\phi$. 

For the forward direction, if $p \Vdash_s \lnot \lnot \phi$ then the set of $r \leq p$ such that $r \Vdash_s \phi$ is dense below $p$. Hence any generic filter containing $p$ will contain such an $r$, and thus $M[G]$ will satisfy $\phi$ by the lemma. 
Conversely, suppose that for every generic $G$ containing $p$, $M[G]$ satisfies $\phi$. For contradiction assume there is a condition $q \leq p$ that forces $\lnot \phi$. Then there is a generic filter $G$ containing $q$ and $p$, and by the lemma we know $M[G] $ satisfies $\lnot \phi$. This is impossible by hypothesis, since $p \in G$. 
Thus by the definition of $\Vdash_s$, for every $q \leq p$ there is some $r \leq q$ with $r \Vdash_s \phi$. This means the set of such $r$ is dense below $p$, which immediately implies that $p \Vdash_s \lnot \lnot \phi$. 

Proposition 2. If $M[G]$ satisfies $\phi \Leftrightarrow \psi$ for every generic $G$ containing $p$, then $p \Vdash_s \lnot\lnot\phi$ if and only if $p \vDash_s \lnot\lnot\psi$. 

Suppose $p \Vdash_s \lnot\lnot\phi$. Then by Proposition 1, $M[G] \vDash \phi$ for every generic $G$ containing $p$. Hence $M[G] \vDash \psi$ for every generic $G$ containing $p$, by hypothesis. Applying the proposition again, $p \Vdash_s \lnot\lnot \psi$. 
