Q:

Understanding power dissipation of a diode in a datasheet chart

According to this chart of a diode's power dissipation over temperature, the power dissipation would decrease as temperature goes up.
But wouldn't it be the exact opposite and even more power be dissipated as heat, when the temperature goes up ?

A:

The graph is not showing how much power the diode will dissipate as a function of temperature.
It shows permissible power dissipation as a function of ambient temperature.  Think about it.  If it's hot all around, less power will get something to a particular temperature than when it's cold all around.
In this case it looks like the diode stops functioning at 175 °C.  If it's already at 174 °C, then you can't put much power into it before it won't function anymore.  If it's at 25 °C, then you can put a lot more power into it before it gets to 175 °C and stops working.

A:

But wouldn't it be the exact opposite and even more power be dissipated as heat, when the temperature goes up?

No. The ability of the diode to dissipate heat will be proportional to \$ \frac {\Delta T}{R_T} \$ where \$ \Delta T \$ is the difference in temperature between the chip and the ambient and \$ R_T \$ is the thermal resistance between the chip and ambient. The latter parameter covers the thermal resistance of the chip to case, case to heatsink (if provided) and heatsink to air or coolant.
Since the maximum temperature of the chip is limited to 175°C then as \$ \Delta T \$ decreases due to increasing ambient temperature the power to be dissipated must decrease to stay in the safe operating area (SOA).
Remember that temperature of the device rises until the power lost to cooling equals the electrical power input. \$ P_{IN} = P_{OUT} \$.

From the comments:

So the watts in the "Power dissipation" is actually current * voltage. 

Yes, but remember it's just the voltage across the diode which, for silicon diodes, will be about 0.7 V. So at 1 A you will have \$ P = VI = 0.7 \times 1 = 0.7 \ \mathrm W \$ dissipated in the diode.

... and the more watts goes through the diode (if i can put it that way), ... 

Probably not the best way. Watts that "go through" the diode won't affect it as they'll get to the load. Watts dissipated at the diode are the problem.

... the faster it heats ...

Yes.

... and so as the ambient temperature goes up the less watts are needed to increase it's temperature to 175C in which the diode would stop functioning? 

Correct.

