Note that he drop the subscript $\langle\rangle_{S^{2}}\rightarrow\langle\rangle$ when he write $\langle (v+q)(\tilde v+\tilde q)\rangle$ so you should not interpret this as a true correlation function. It is just a notation tool that means: sum over all contractions of $q's$ using $-\eta^{\mu\nu}(z-z')^{-2}\alpha '/2$ so
$$ \langle \partial x^{\mu}(z) \partial x^{\nu}(z')\rangle_{S^{2}} \neq \langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle $$
The equation above becomes an equality only if we set $v^{\mu}(z)=v^{\nu}(z')=0$, i.e. if there is no exponential.
Let us consider the case with two $\partial x$'s and various exponentials. The expectation value becomes
$$ iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times \langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle $$
where
$$ \langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle = v^{\mu}(z)v^{\nu}(z')+v^{\mu}(z)\langle q^{\nu}(z')\rangle+\langle q^{\mu}(z)\rangle v^{\nu}(z')+\langle q^{\mu}(z)q^{\nu}(z')\rangle $$
and $\langle q^{\nu}(z')\rangle=\langle q^{\mu}(z)\rangle=0$ while
$$ \langle q^{\mu}(z)q^{\nu}(z')\rangle = -\eta^{\mu\nu}\frac{\alpha '}{2}\frac{1}{(z-z')^{2}} $$
so the effect of replacing $\partial x$ by $(v+q)$, and treating $q$ as $\partial x$, i.e. contracting $q$'s as if they are $\partial x$'s, has the effect of taking account for the contractions of $\partial x$'s with the exponentials. The reason why this works is because the exponential is a kind of "eigenvector" under contractions with $\partial x$, where the "eigenvalues" are the $v$'s.
Another archetype is the correlation between one $\partial x$ and various exponentials. Using the formula proposed by Polchinski this gives
$$ iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times \langle (v^{\mu}(z)+q^{\mu}(z))\rangle = $$
where now
$$ \langle (v^{\mu}(z)+q^{\mu}(z))\rangle = v^{\mu}(z)+\langle q^{\mu}(z)\rangle = v^{\mu}(z) $$
which confirms that $v^{\mu}(z)$ works as a "eigenvalue" and the exponentials as "eigenvector" of $\partial x^{\mu}(z)$. There is a physics behind this! The $\partial x^{\mu}(z)$ is a conserved current associated to momentum + winding, while the exponentials are, via state-operator correspondence, associated to states with well defined momentum and winding number.
The contraction of a current with a local operator is related, again by state-operator correspondence, to the conserved charge operator acting on the state.