Why does Carter-Wegman MAC have 2n-bit output?
The MAC consists of the pair of the n-bit random value $r$ and the n-bit value $t = F(k_1,r) \oplus S(k_2,m)$. The length of $(r,t)$ is at least $2n$ bits, partly depending on how it is encoded. If the two values are simply concatenated, the total length is $2n$, as stipulated.