Let't take a simple process for example $$e^-+e^+ \to \mu^-+\mu^+$$ The only tree level diagram for this process is the following
To build up the matrix element of this process you just have to follow the fermion lines backwards, that's all. Beginning from the end of the process you have then $$(\mu^- \to \text{vertex}\to\mu^+)\text{ Photon propagator }(e^+\to\text{vertex}\to e^-)$$ which gives $$\mathcal{M}=\bar u(p_3,\sigma_3)(-ie\gamma^\nu)v(p_4\sigma_4)\frac{-ig_{\mu\nu}}{q^2+i\epsilon}\bar v(p_2,\sigma_2)(-ie\gamma^\mu)u(p_1,\sigma_1)$$ with $q=p_1+p_2$
This holds for every Feynman diagram in general, you have to follow backwards the fermion lines and use the appropriate spinors for particles and antiparticles. The convention of starting from the end does not matter, you could as well start from the beginning as long as you follow the fermion lines backwards.
Edit
Since the OP asked for the case of Compton scattering I add it. So the s-channel Compton scattering process is given by this diagram (and the one with the photons with exchanged positions) 
where the two fermion lines are just electrons. As we said before let's follow the fermion lines backwards $$(e^-\to\text{vertex}\to\text{photon})\text{Fermion propagator}(\text{photon}\to\text{vertex}\to e^-)$$ which gives
$$\bar u(p^\prime, \sigma^\prime)(-ie\gamma^\nu)\epsilon_\nu^{*}(k^\prime)\frac{\not p+\not k+m}{(p+k)^2-m^2+i\epsilon}\epsilon_\mu(k)(-ie\gamma^\mu)u(p, \sigma)$$