The electron exchanges a virtual photon with the positron without annihilating. The photon carries momentum $k = p - p'$ with $k^2 = t$. Dominates at small scattering angles (forward scattering).
The computational chain
From diagram to number—every step of the amplitude calculation.
The diagram
The electron and positron exchange a virtual photon without annihilating. The photon carries momentum $k = p - p'$ with $k^2 = t$.
Feynman rules expression
$$i\mathcal{M}_t = \frac{ie^2}{t}\left[\bar{u}(p')\gamma^\mu u(p)\right]\left[\bar{v}(q)\gamma_\mu v(q')\right]$$
The electron current $\bar{u}\gamma^\mu u$ and the positron current $\bar{v}\gamma_\mu v$ are connected by the photon propagator.
Square and sum over spins
$$\overline{|\mathcal{M}_t|^2} = \frac{1}{4}\frac{e^4}{t^2}\,\mathrm{Tr}\left[(\not{p}'+m)\gamma^\mu(\not{p}+m)\gamma^\nu\right]\,\mathrm{Tr}\left[(\not{q}-m)\gamma_\mu(\not{q}'-m)\gamma_\nu\right]$$
The electron completeness gives $(\not{p}+m)$; the positron completeness gives $(\not{q}-m)$ (note the minus sign for $v$ spinors). Each bracket becomes a trace.
Evaluate the traces
$$T_1^{\mu\nu} = 4\left(p'^\mu p^\nu + p^\mu p'^\nu - g^{\mu\nu}(p\cdot p' - m^2)\right)$$
$$T_2^{\mu\nu} = 4\left(q^\mu q'^\nu + q'^\mu q^\nu - g^{\mu\nu}(q\cdot q' - m^2)\right)$$
Both completeness relations have the same sign on $m$: $(+m)(+m)$ for the electron trace and $(-m)(-m)$ for the positron trace. In both cases $m^2$ enters with a positive sign, giving $-(p\cdot p' - m^2)$ in the metric term. The cross terms vanish by the odd-trace rule.
Contract and simplify (massless limit)
$$\overline{|\mathcal{M}_t|^2} = \frac{2e^4(s^2+u^2)}{t^2}$$
Contract $T_1^{\mu\nu}T_{2\mu\nu}$ and substitute Mandelstam variables in the massless limit. The $1/t^2$ pole gives the characteristic forward peak of Coulomb scattering.