Let $V$ be a finite dimensional space over the field $\mathbb{F}$ with inner product $\langle \cdot, \cdot \rangle$. Then for every linear form $f: V \rightarrow \mathbb{F}$ there exists a unique $z \in V$ such that $f(w)=\langle w,z \rangle$.
$V$ and its dual space $V^*$ are isomorphic because they have the same dimension and they are over the same field.
One such isomorphism is $\psi : V \to V^*$ where $\psi : v \mapsto \langle v, \cdot \rangle $.
Then $\psi(v)(w)=\langle v,w \rangle $ and the fact that $\psi$ is an isomorphism tells us that there indeed exists only one vector $v$ for which the form is expressed as $\langle v, \cdot \rangle$.
Is that correct?