DC voltage after rectifier
It was written: "6 * 1.414 = 8.484 VacRMS No load factor 8.484 * 1.64 = 13.91 After rectifier 13.91 - 1.4 = 12.51 Vdc"

Because it is a 'short-proof' transformer, it will be inefficient. 

I interpreted the calculation this way and presumed to fix the apparent transposition of 1.414 and 1.64. I hope it makes sense:

6 * 1.64 No load factor = 9.84 VacRMS (no load)
(means 6VAC at full load 2VA/333mA, and 9.84V at no load)

9.84 VacRMS * 1.414 = 13.91VAC peak (no load)

13.91 - 1.4 = 12.51VDC after rectifier and cap filter (no load).

So that is no-load. 
Now for 300mA load:

This is rough approximations because no curve is published for transformer:
no load - full load = voltage drop
9.84VAC - 6VAC = 3.84VAC
0mA to 333mA = drop of 3.84VAC

volts drop per mA = 0.01153 (if linear)

0.01153V/mA * 300mA = 3.45945VAC drop from 9.84VAC

9.84V - 3.45945V - 6.380V
For the 300mA load, VAC is 6.380

It will probably look like this:
6.380V * 1.414 = 9.0221V peak
9.02V - 1.2V bridge rectifier drop = 7.8221VDC peaks to charge filter.
Filter charged to 7.8V
LDO needs at least 6.2VDC at 800mA

So even with some mistakes, there should be enough voltage for your 1.2V dropout 5V regulator at 300mA. 

If there is still too much doubt, suggest either ordering the transformer and just trying it on the workbench, or, contacting the manufacturer or distributor application engineer, sending your schematic over and asking them if it's OK.

The dissipation and very small size of the transformer also suggests using large copper tracks for each transformer lead to help carry away the heat.