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.
[Edit] 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. [/Edit]