How do you decide if a process has $s$ or $t$ channel Feynman diagram?

You can determine this by using the vertices of the Standard Model (below) and building a Feynman diagram out of these vertices.

The interaction $rm e^+e^-to e^+e^-$, known as Bhabha scattering, has interfering diagrams in both the s- and t-channels. In the s-channel, the two particles annihilate into a photon that then turns back into an electron and a positron, while in the t-channel, the two particles exchange an electron. Both of these processes use the second vertex below twice.

$rm e^+e^-togammagamma$, on the other hand, has no s-channel. This is because the positron and the electron can annihilate into a photon (or a Z boson), but neither a photon nor a Z boson couples to two photons. That is, there is no 3 photon vertex or $rm Zgammagamma$ vertex below. The only way to form $rm e^+e^-togammagamma$ at tree level is with an electron as an internal line.