How do I calculate the amplitude of an $s$-channel Feynman diagram? What is is the starting point to measure the amplitude in an $s$-diagram like the one below where two particles collide and create a propagator followed by the final product?
I know that in $t$-diagram I should start from the opposite side to the direction of propagation but in this case, the particles are propagating to the same point, meaning:

What I mean is:
Should I start by adding the maths expressions from the top right to bottom left, as in $Z_L(p_3)$, followed by the vertex and then by the mathematical expression for $Z_L(p_4)$ followed by the propagator?
 A: As the diagram is rather general, i.e. it is unknown if the particles $Z_L(p_i)$ are distinguishable or undistinguishable, particles or anti-particles, bosons or fermions, one can only make very general statement. Anyway, it is possible to start translating the Feynman-diagram wherever you want, as composing the mathematical expression is essentially multiplication and multiplication is commutative.
The expression for the scattering amplitude would be in the most general form ($g_i$ with $i=1,2$ are the coupling constants which are not necessarily the same). I assume a scalar coupling:
$$i{\cal M} = J(1,2)\frac{-i g_1 g_2}{s -m^2_{H_0}}J(3,4)$$
where $J(1,2)$ is the "current" (caveat: possibly this "current" is not conserved, here this does not matter) of the particles $Z_L(p_1)$ and $Z_L(p_2)$, and $J(3,4)$ is the "current" of $Z_L(p_3)$ and $Z_L(p_4)$. If the particles are not distinguishable, another diagram has to be added where the outgoing particles are swapped with respect to the ingoing particles. 
BONUS:
As this diagram is supposed to be s-channel process, $Z_L(p_1)$ and $Z_L(p_2)$ would annihilate, and $Z_L(p_3)$ and $Z_L(p_4)$ would be created. So in case of fermions, the "currents" would be something like:
$$J(1,2) = \overline{v}(-p_2)u(p_1)  \quad \text{and} \quad J(3,4) = \overline{u}(p_4)v(-p_3)$$
but I don't guarantee that this expression is 100% correct, it is just for giving you an idea. 
EDIT:
Actually, the products of the bispinors, for instance $\overline{u}\cdot v$, are not commutative, so one could define a rule in which order they should be written up (Nevertheless $J(1,2)$ and $J(3,4)$ can be still commutated). However, it would be only valid for a $s$-channel process, for the other channels other rules would have to be applied. 
ANOTHER EDIT:
The signs of the momenta depend on the assumptions that $p_1$ corresponds an in-going particle and $p_2$ corresponds to an outgoing particle. Finally, $p_3$ is considered as an in-going particle and $p_4$ corresponds to an out-going particle. If the direction of a particle is swapped, the sign has to be swapped correspondingly. 
