Case 1: 2 tickets, one lottery, one prize, $n$ different tickets equally likely
Events: A:= win with ticket A, B:= win with ticket B, W:= win the prize
$P(W)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=p + p - 0 = 2p$
Notice that $P(A\cap B)=0$ (you can't win with 2 tickets in the same lottery), events $A$ and $B$ are not independent, they are mutually exclusive.
Case 2: 2 lotteries, one ticket in each lottery (2 possible prizes), $n$ tickets equally likely in each lottery
Events: A:= win in lottery 1, B:= win in lottery 2, W:= win in at least 1 lottery,
Z:=win in exactly one lottery
$P(W)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=p + p - p^2 = 2p-p^2$
The events $A$ and $B$ are independent in this case (not in Case 1), therefore $P(A\cap B)=P(A)P(B)=p^2$
If you want the probability of winning in exactly one lottery (event $Z$) but not in both:
$P(Z)=P((A\cap \bar B)\cup(\bar A \cap B))=P(A\cap \bar B)+P(\bar A\cap B)-P(\varnothing)$
Conclusion: the probability of winning in Case 2 (in at least a lottery) is a bit smaller than in Case 1 as $p^2=1/n^2>0$ is a very small positive number ($\rightarrow 0$ as $n\rightarrow \infty$), but you get as a bonus the possibility of winning 2 prizes if you are very lucky! not possible in Case 1 :).