It is not necessary, as such, but it provides a way of seeing how the dot product form arises.
The relevant fact is that any linear functional $f: V \rightarrow \unicode{x211D}$ can be written in terms of the dot product, ie, $\exists \phi \in V$ such that $f(v) = v . \phi$. Furthermore, this $\phi$ is unique (this is important).
Now pick a $w \in V$, and consider the (single-variable) functional $u \mapsto \tau(u, w)$. Since it is linear, there exists an element of $V$ so that it can be written in terms of the dot product. Let me label this element $\phi_\tau(w)$. Then we have $\tau(u, w) = u . \phi_\tau(w)$. To finish the proof, we just need to show that the function $\phi_\tau: V \rightarrow V$ is linear.
Look at $\tau(u, \lambda w)$, for some scalar $\lambda$, then we have $\tau(u, \lambda w) = u . \phi_\tau(\lambda w)$, but since $\tau$ is linear in the second variable, it is also true that $\tau(u, \lambda w) = \lambda \tau(u, w) = u . (\lambda \phi_\tau(w))$. By uniqueness, we conclude that $\phi_\tau(\lambda w) = \lambda \phi_\tau(w)$.
In a similar manner, consider $\tau(u, w_1+w_2)$, with $w_1,w_2 \in V$. Along previous lines, we have $\tau(u, w_1+w_2) = u.\phi_\tau(w_1+w_2)$. Also $\tau(u, w_1+w_2)= u. (\phi_\tau(w_1) + \phi_\tau(w_2))$, so by uniqueness we have $\phi_\tau(w_1+w_2) = \phi_\tau(w_1) + \phi_\tau(w_2)$, and so $\phi_\tau$ is linear.
Since $\phi_\tau$ is linear, it can be written in the form $\phi_\tau(w) = T w$, where $T$ is a linear operator. Hence the expression $\tau(u, w) = u . (Tw)$.