This is to a great extent educated guessing, as the Wikipedia page was kinda terse on the definition department :-)
Anyway I think I have an explanation to your first question. As the abstract oval is not (originally) defined by a polynomial equation, it is kind of difficult to come up with a definition of a tangent. The following seems to fit the bill (correct me if I'm wrong). Fix a point $X$ on the oval. There are $q+1$ lines going thru it. There are $q$ other points on the oval. Each and every one of them belongs to exactly one of the $q+1$ lines. Furthermore, no two points of the oval (other than $X$) belong to the same line, as that would violate the assumption that no three points are collinear. Therefore the $q$ non-$X$ points of the oval are on $q$ distinct lines thru $X$. So there is exactly one line thru $X$ that does not intersect the oval elsewhere. I assume that this line is referred to as the tangent of the oval at $X$.
On with the first question. With the coordinate system now set up (we are free to do it in this way, because the group of projective linear transformations is 3-transitive), we see that the $q+1$ lines thru the point $P$ are the lines $Z=0$ (1 choice) and $Y=aZ$ ($q$ choices, one for each value of the constant $a$). Because the line $Z=0$ goes thru $Q$ also, it cannot be the tangent of the oval at $P$. Therefore the tangent is of the other form, for some $a\in GF(q)$. [Edit] Because the point $R$ cannot lie on the tangent either, we can say that $a\neq0$. [/Edit] I dare guess that's all there is to your question 1), but it is hard to tell without access to the textbook.
Question 2) is harder. Kinda makes sense, if the Wikipedia article mentions the list as a published paper. It is obviously true in the Fano plane ($q=2$), where our answer to question 1) tells us that the tangent lines must be $Y=Z$, $X=Y$ and $X=Z$ and they all meet at the point with projective coordinates $(1:1:1)$. A curious result anyway.