Specific current mirror explanation-source
Here is how I understand this current mirror:

The upper two transistors are naturally designed to operate in the saturation region since the upper left transistor is diode connected (the upper right is not diode connected, but assuming there's a high enough voltage on its drain terminal, it will always be in saturation due to sharing a gate connection with the upper left transistor). 

But the lower two transistors must be operating in triode. This occurs because the gates of the lower two transistors are pulled up to the same voltage on the gates of the upper two transistors (unlike in a normal cascode configuration). Thus, the overdrive voltage (\$V_{GS} - V_{th}\$) of the lower two transistors is higher than the upper two. Now note that all 4 transistors are running the same drain current, and they are all ideally matched. A higher overdrive voltage on the lower transistors will make them, should they operate in saturation, want to pass more current than the upper transistors. That is clearly impossible, so they must be operating in saturation.  

Since the bottom two transistors are operating in saturation they look like the upper transistors, like the \$R_{DSon}\$ of the transistors. This has a similar effect as the degeneration resistor in a common-source amplifier, and increases the effective output resistance of the current mirror. Now because the \$R_{DSon}\$ of a MOSFET is usually in the \$m\Omega\$ region, this degeneration does virtually nothing. 

A proper cascode circuit uses a similar configuration to achieve a similar effect, except the upper right transistor would see roughly the \$r_o\$ of the lower transistor instead of the \$R_{DSon}\$ -- a much greater quantity (many orders of magnitude).

I ran a couple SPICE simulations in LTSPICE to demonstrate the effect. 

Here's the schematic. On the left is the standard current mirror circuit. The middle is the "cascode" described in the patent. The right is the textbook cascode circuit. The reference current is set to 1mA, and the mirror is sinking that 1mA into a 5k load.



Now here's the results:



The blue line is the ordinary current mirror, the green line in the patent current mirror, and the red line is the cascode current mirror. The current accuracy of the patent current mirror is significantly worse compared compared to the classic cascode mirror at this output voltage. In this case, the patent current mirror is even worse than the classic current mirror, but this may not be true with all transistors and at all output voltages. A cascode mirror is the clear winner here. 

The poor performance of the patent mirror is, I believe, due to the voltage at the drain of the upper left transistor being different to the top right. Thus the bottom left transistor will see a different drain voltage than the bottom right transistor. This is seen in the simulation. This is, of course, the sort of behavior a cascode mirror is designed to rectify.

Lastly to demonstrate that the lower transistors of the patent cascode are in triode, here's the voltages from drain to source:



So their \$V_{DS}\$ is on the order of millivolts -- what you'd expect for a triode MOSFET. \$V_{DS}\$ is much smaller than \$V_{GS} - V_{TH}\$, so they have to be in triode.