The in and out states you use have two particles in them. The operator you mention with the double contraction only has a single annihilation and creation operator in it, so it only acts on a single particle from the two particle states. There is no momentum transfer due to this operator. Momentum conservation then constrains the outgoing momenta to equal exactly the incoming momenta and you have forward scattering, i.e. no scattering.
I did a similar thing to you and tried the meson-nucleon scattering. The Hamiltonian is $H_{int}=\int d^3x g\psi^{\dagger}\psi\phi$ and we want to find $<f|S|i>$ to second order in $g$. So we're interested in the time-ordered bit:
$$T\left[\psi^{\dagger}(x_1)\psi(x_1)\phi(x_1)\psi^{\dagger}(x_2)\psi(x_2)\phi(x_2)\right]$$
Applying Wick's Theorem, the relevant parts are:
$$
:\psi^{\dagger}(x_1)\phi(x_1)\psi(x_2)\phi(x_2):\overbrace{\psi(x_1)\psi^{\dagger}(x_2)}+\\
:\psi^{\dagger}(x_1)\phi(x_1)\psi(x_2)\phi(x_2):\overbrace{\psi^{\dagger}(x_1)\psi(x_2)}
$$
These two are equivalent. If you solve the top one say (they are already normal ordered in their form above), you get
$$
<f|S|i>=\frac{(-ig)^2}{2}(2\pi)^4\delta^{(4)}(p_1-p_1'+p_2-p_2')\frac{i}{(p_1+q_1)^2-M^2+i\epsilon}
$$
Solving the other one gives the same result so, adding them together gets the thing above without the factor of $1/2$.
Now, if you normal order them the second way as you suggested i.e.
$$
:\psi^{\dagger}(x_1)\phi(x_2)\psi(x_2)\phi(x_1):\overbrace{\psi(x_1)\psi^{\dagger}(x_2)}+\\
:\psi^{\dagger}(x_1)\phi(x_2)\psi(x_2)\phi(x_1):\overbrace{\psi^{\dagger}(x_1)\psi(x_2)}
$$
which is equally plausible given $\phi^{\dagger}=\phi$, you would get a final amplitude of
$$
<f|S|i>=(-ig)^2(2\pi)^4\delta^{(4)}(p_1-p_1'+p_2-p_2')\frac{i}{(p_1-p_1')^2-M^2+i\epsilon}
$$
This total amplitude is what you'd get as a result of drawing the Feynman diagrams. I suppose the result of this is that Wick's theorem needs to consider $\mathit{all}$ normal orderings in order to get a complete picture!
Best Answer
By considering matrix elements of the form $\langle \vec{q} | S | \vec{q}' \rangle$, Coleman is considering processes in which a single meson enters and a single meson exits. You're completely correct that there are pair production processes; but the initial and final states must contain single mesons.
If you were to calculate the $S$ matrix elements listed, you would consider Feynman diagrams containing intermediate pair production and annihilation processes. All of these will contribute to the self energy of the particle, which would renormalize the mass in the absence of counterterms. Coleman is saying that the method to determine the counterterm coefficients $b$ and $c$ is to compute the mass of the particle, including the self-energy corrections and counterterms, and demand that they are equal to the physical masses.