Kaarel wins. Suppose Richard chooses $a$ in the beginning. Kaarel then chooses $p-a$. Next, Richard chooses $b$. If Kaarel's number, $c$, satisfies: \[ a(p-a)+bc\equiv 0\pmod{p}\Leftrightarrow c\equiv \frac{a^2}{b}\pmod{p}, \]Kaarel wins. Let us now ensure that $c\triangleq \frac{a^2}{b}$ is not chosen before. Since $a\not\equiv b\pmod{p}$, $c$ is distinct from $a$. Likewise, if $c\equiv -a\pmod{p}$, then $b\equiv-a\pmod{p}$, but $p-a$ is already chosen by Kaarel. Finally, if $c\equiv b\pmod{p}$, then $a\equiv \pm b\pmod{p}$. But both $a$ as well as $p-a$ are chosen in the beginning respectively by Richard and Kaarel. Hence, Richard cannot choose one of these in round 2 as his $b$. Hence the conclusion.

Can't we just do this? If Richard chooses $r$, Kaarel can choose $p-r$. So the sum after the final move is conguruent to $- \left( 1^2 + 2^2 + \cdots \left( \frac{p-1}{2} \right)^2 \right) \equiv - \left( \frac{\frac{p-1}{2} \cdot \frac{p+1}{2} \cdot p }{6} \right) \equiv 0 \pmod{p}$ so we are done and Kaarel must win?