Riesz representation theorem giving a different result?

Evans p298 - Lax Milgram Theorem

Let $H$ be a Hilbert space. Denote the inner product by $(\cdot,\cdot)$ and the natural dual pairing of spaces by $\langle\cdot,\cdot\rangle$.

He gives us bilinear form $B[\cdot,\cdot]:H\times H \to \Bbb R$ and then he says that for each fixed element $u\in H$ the mapping $v\mapsto B[u,v]$ is a bounded linear functional on $H$, and hence Riesz representation theorem gives us a unique element $w\in H$ such that: $$B[u,v]=(w,v),\qquad (v\in H)$$

How exactly does this follow from RRT? Riesz gives us that any time we consider $u^*\in H^*$ there is some $u\in H$ such that: $$\langle u^*,v\rangle = (u,v).$$

So I don't follow how the RRT gives the above, I mean unless the RRT doesn't hold just for the natural pairing, but rather for any bilinear form?

Well, $v \to B[u,v]$ is a linear functional on $H$. The RRT tells you that any bounded linear functional $l$ on $H$ can be represented as $l(v)=(u,v)$ for some $v$, and that the map $l \to u$ is in fact an isometric isomorphism. That is why you usually write $l=u^*$. Hence, since $v \to B[u,v]$ is a bounded linear functional you will find a $w$ such that $B[u,v]=(w,v)$, you may write $B[u,\cdot]=w^*$ if you like.