Group Signatures with VLR Revocation Check
I do not have access to the paper you have linked, but I certanly can answer the first question and maybe answer the third question:


$(\hat{u}, \hat{v}) \in G_2^2$ is a short version of $(\hat{u}, \hat{v}) \in (G_2 \times G_2)$ which is a short version of "$\hat{u} \in G_2$ and $\hat{v} \in G_2$".
Since I don't have access to the paper I'm gonna make some assumptions to answer the third question:


$e$ is a bi-linear mapping of the form $G_1 \times G_2 \rightarrow G_3$. I make this assumption based on the fact the second parameter is element of a group called $G_2$ which hints at the existence of a group $G_1$.
$\psi$ is a linear homomorphism from $G_2$ to $G_1$. This would make sense as the first parameter of $e$ would be an element of $G_1$ under the first assumption. Also $\psi(g_2)$ must be $g_1$ for the math below to check out.



Be $x, y \in Z_q$ with $\hat{u} = g_2^x$ and $\hat{v} = g_2^y$, then $u = g_1^x$ and $v = g_1^y$ since $\psi(g_2) = g_1$. This leads to:

$e(v^{\alpha}, \hat{u}) = e((g_1^y)^{\alpha}, g_2)^x = e(g_1^{\alpha y}, g_2)^x = e(g_1, g_2)^{(\alpha y) x} = e(g_1, g_2)^{(\alpha x) y} = e(g_1 ^ {\alpha x}, g_2^y) = e((g_1^x)^{\alpha}, g_2^y) = e(u^{\alpha}, \hat{v})$

AS stated above, this only works under the assumptions above, which are assumptions. If these are not correct feel free to leave a comment or edit.