What does zero transfer function mean?
You have a system where not just some states, but the output itself is not controllable or observable. There are a couple of ways to look at what is happening.

Approach A 

The third state equation is \$x_3'=-x_3\$. Since the conversion to transfer function assumes all initial conditions are zero, the solution for \$x_3\$ is \$x_3=0\$.

Now we are left with two equations: \$x_1'=2 x_1-x_2+u\$ and \$x_2'=x_2+u\$

So \$x_1'-x_2'=(2 x_1-x_2+u)-(x_2+u)= 2 (x_1- x_2)\$

The output is \$y = x_1- x_2 \$ and its equation becomes \$y'=2 y\$, whose solution is also 0. 

Approach B

Another way of looking at it is to convert the model to Jordan form. 


Now the first state equation is \$z_1'=-z_1\$ and the last one is \$z_3'=2 z_3\$. Again the solutions for these two equations are \$z_1=z_3=0\$.

The output in terms of the Jordan states is \$y=-4 z_1+ z_3=0\$. 

The output is just a linear combination of the two uncontrollable states. The other state \$z_2\$ is controllable but it is not observable and does not show up in the output.