Multiplication of polynomials in value representation as done for Fast Fourier Transform
If $P, Q, R \in \mathbb{R}[X], R = PQ$, then for any $x \in \mathbb{R}$, $R(x) = P(x)\times Q(x)$.
It means that if the value representation of $P$ is $[(x_1, P(x_1)); (x_2, P(x_2)); …, (x_n, P(x_n))]$ and the value representation of $Q$ is $[(x_1, Q(x_1)); (x_2, Q(x_2)); …, (x_n, Q(x_n))]$, then the value representation of $R$ is $[(x_1, P(x_1)Q(x_1)); (x_2, P(x_2)Q(x_2)); …, (x_n, P(x_n)Q(x_n))]$.
In your example, $(-2, 1)$ represents a couple $(x_i, P(x_i))$ and $(-2, 9)$ is $(x_i, Q(x_i))$, so it's normal that $(x_i, R(x_i))$ is $(-2, 1\times 9) = (-2, 9)$.