Why doesn't this put $BPP$ in $NP$?
Adleman's theorem states that you can amplify the success probability of $M$ so that there will be a single choice for the random string $r$ that works for all $x$:
$$ \exists r \forall x : x \in L(M) \Leftrightarrow M(x,r) = 1. $$
(Here $x$ goes over all inputs of length $n$, and $r$ goes over all possible random strings that $M$ requires for inputs of length $n$.)

Unfortunately, it's not clear how to find $r$, and even worse, if I give you $r$, it's not clear how to verify that it satisfies the property stated above. This is why we have to supply it as advice.

The Sipser&ndash;G&aacute;cs theorem similarly states a different property from what you list. It states that if $M$ is a BPP machine then
$$
\forall x \bigl(x \in L(M) \Longleftrightarrow \exists t_1,\ldots,t_{|r|} \forall r : M(x,r \oplus t_1) = 1 \lor \cdots \lor M(x,r \oplus t_{|r|}) = 1 \bigr).
$$
Here the strings $t_i$ could depend on the input $x$. This is why you get $\Sigma_2^P$ or $\Pi_2^P$ rather than $\mathsf{NP}/\mathsf{poly}$ or $\mathsf{coNP}/\mathsf{poly}$. 



Summarizing, you have to be careful with quantifiers when stating a theorem. Changing the order of the quantifiers could drastically change the meaning.