Ok, there are many things at play here. Let me just quickly define a few things for you. I'm assuming some background knowledge since you mentioned the Shockley equation.
A diode is formed by joining a piece of n-type and p-type semiconductor. This leads to diffusion of electrons and holes which creates a current. As a result, the space charge region (SCR) is formed. The space charge region creates an electric field that creates a drift current that cancels off the diffusion current. Hence at thermal equilibrium, there is no current.
A zener diode depends on quantum tunneling. This means that the breakdown voltage is achieved once the p-region valence band edge is raised above the n-region conduction band edge. This allows the electrons in the p-type valence band to tunnel to the conduction band of the n-type region. This creates a current.
An avalanche diode (that's not a regular diode, its an avalanche diode), depends on the avalanche effect. When the SCR field exceeds a certain amount (known as critical field), electrons get accelerated to very high speeds and start knocking other electrons into the conduction band. This creates a huge current. Notice the difference in operating principle between the zener and the avalanche diodes.
Ok, now to tackle the questions.
This analysis is simplified but should be good enough. In a regular diode, when you raise the temperature, the carrier concentrations rise greatly. This affects diffusion current only minimally as the rise is around the same on both sides so we can approximate diffusion current to be constant for small increases in temperature. Drift current increases proportional to the carrier concentrations however and so drift current increases greatly. This means that a smaller electric field is required in the SCR to offset the diffusion current. Due to this smaller electric field, the turn-on voltage of the diode decreases.
In a zener diode, when you raise the temperature, the energy of electrons increases. Consequently, the tunnelling probability increases and the reverse breakdown voltage drops. (not entirely sure of this but seems plausible)
In an avalanche diode, when the temperature is higher, the built in field drops as per the previous explanation. Hence, a larger applied voltage is needed to reach the critical field and so breakdown voltage increases.