Can I convert $F_{q^{12}}$ to $F_q$?
The paper writes $z^r \cdot G$; however $z^r$ is a member of the extension group $\mathbb{F}_{q^{12}}$, while point multiplication is formally defined over the integers; you ask &quot;what are we supposed to do here?&quot;
Well, going through the paper, it appears that if we rewrite that equation to $h(z^r) \cdot G$, where $h$ is a function from $\mathbb{F}_{q^{12}}$ to $\mathbb{F}_{q}$, that works (assuming, of course, we rewrite the decryption process similarly), so we have:
$\text{Encrypt}(pk, m) = (r \cdot pk, h(z^r) \cdot G + Pm)$
$\text{Decrypt}^1(C, sk) = B - h( e( A, sk^{-1}G )) \cdot G$
$\text{Decrypt}^2(C, sk) = B - h( A^{1/b} ) \cdot G$
(see the paper for explanation of the various notations, and the reencrypt process doesn't change)
Any deterministic $h$ would work (in the sense that the protocol will work), as can be seen by going through encryption/reencryption/decryption steps.  My inclination would be to use a hash function.