Yes, you can do this using polar decomposition. We can also consider $\phi$ to be a normal linear functional on $A^{**}$, and there is a positive $\omega \in A^*$ and a partial isometry $v \in A^{**}$ such that $\phi(a) = \omega(va)$ for all $x \in A$. (I'm sure this is in volume 1 of Takesaki, probably also in Pedersen.) We have $\|\omega\| = \|\phi\| \leq 1$, so we can apply GNS to $\omega$ and get $\phi(a) = \omega(va) = \langle \pi(va)\xi,\xi\rangle = \langle \pi(a)\psi,\eta\rangle$ with $\psi = \xi$ and $\eta = \pi(v^*)\xi$. As you note, $\|\xi\|^2 = \langle \pi(1)\xi,\xi\rangle = \omega(\xi) = \leq 1$.
Yes, you can do this using polar decomposition. We can also consider $\phi$ to be a normal linear functional on $A^{**}$, and there is a positive $\omega \in A^*$ and a partial isometry $v \in A^{**}$ such that $\phi(x) \phi(a) = \omega(vx)$omega(va)$ for all $x \in A$. (I'm sure this is in volume 1 of Takesaki, probably also in Pedersen.) We have $\|\omega\| = \|\phi\| \leq 1$, so we can apply GNS to $\omega$as you say the vector\omega$ and get $\xi$ achieves the conclusion for \phi(a) = \omega(va) = \langle \pi(va)\xi,\xi\rangle = \langle \pi(a)\psi,\eta\rangle$ with $\omega$\psi = \xi$ and $\eta = \pi(v^*)\xi$. As you note, $\|\xi\|^2 = \langle \pi(1)\xi,\xi\rangle = \omega(\xi) = 1$`.