Riesz representation theorem says that if $H$ is a Hilbert space, and $f\in H^*$, then there exists a unique $y \in H$ such that $f(x)=\langle y,x \rangle_H$.
Now suppose we look at $H\times H$. This is still a Hilbert space, equipped with the scalar product $\langle (x_1,x_2),(y_1,y_2)\rangle_{H \times H}= \langle x_1,y_1 \rangle_H + \langle x_2,y_2 \rangle_H$
Clearly we can apply Riesz representation theorem to $H\times H$: $\forall f \in (H\times H)^*$, there exist a unique $(y_1,y_2) \in H \times H$ such that $f(x_1,x_2)= \langle (y_1,y_2), (x_1,x_2)\rangle_{H\times H}$
The (unproved) result in my book says : by Riesz representation theorem, $\forall f :H\times H \rightarrow \mathbb R $ bilinear and continuous, there exist a unique $Q \in L(H)$ such that $f(x_1,x_2)= \langle Qx_1, x_2\rangle_{H}$, where $L(H)$ is the set of bounded linear operators $T:H \rightarrow H$.
How do we prove that ? Do we have to use Riesz on $H\times H$ or simply on $H$ ?
This is not true as stated: the function $(x_1,x_2) \mapsto \langle Qx_1,x_2\rangle$ is not a linear mapping from $H\times H$ to $\mathbb R$. Rather it is bilinear. Please check the assumptions on $f$ again.