Electrical – Current Rating of a device (how does this change when changing the voltage)

The voltage rating is related to the breakdown voltage of the plastic between the pins. You shouldn't exceed the breakdown voltage even of theres no current at all.

The current rating is, as you say, relates to the pin resistance and how much the connector will heat up. You shouldn't exceed the current rating even at very low voltages.

Edit:

As KellenJB says in another answer, it looks like the key thing you're missing is that the power consumed in the connector (and so the self-heating which could damage the connector) is not related to the voltage between the pins, but to the current through the pin. This current, combined with the (very small) resistance of the pin or contact, generates a small voltage between one end of the pin and the other (or between one pin and the socket its mated to). This voltage, multiplied by the current, gives the heat generated in the connector.

Make sure your fuse is rated to break 1.05kVAC. I would suggest fusing the primary as well, since the transformers could fail, in fact it's more important than fusing the secondary.

According to the Hammond Design Guide for Rectifier Use, the maximum DC current output for your configuration is about 0.62 of the AC rating, or about 188mA in your case. This is because the RMS current in the transformer winding (and thus the heating) will be higher than the DC current output.

If you're talking about current available at the terminals as shown, you have to subtract the current through the 400K bleeder resistor, which is about 3.8mA, leaving you with ~184mA.