To use the term "transfer function" is a little ambiguous here. Normally, given an opamp circuit like this, one would want the ratio Vout/Vin, and would call it a "transfer function".
In this case, if you apply a voltage source, Vin, at the node In, the feedback will be shorted to ground by the zero output impedance of a voltage source, and the output voltage of opamp U2 will just be A2*Vin, where A2 is the open loop voltage gain in U2.
What does make sense, however, is to assume that the input is a current. Then the ratio Vout/Iin is what one would like to calculate.
If we calculate the input input impedance, Zin, at V1+, then we can derive:
Vout/Iin = A2*Zin, as A2 and A1 approach infinity.
We can use V1+, V2+ and V2- as nodes with the nodal method of circuit analysis. Denote the voltages at those nodes as V1, V2 and V3. To find Zin, we can inject a current of 1 amp at node A1+; then the voltage there will be equal to the impedance at that node. Setting the open loop gains of both opamps to A, and summing currents at those nodes, we get three equations.
(1/R3 + (1-A)/R4)*V1 + (-A/R3)*V2 + A/R3*V3 = 1
0*V1 + (1/R2 + (1-A)*s*C2)*V2 + (A*s*C2)*V3 = 0
(-A/R1)*V1 + 0*V2 + (1/R1 + s*C1)*V3 = 0
I left some terms multiplied by zero for clarity.
Solve this system and multiply the result for V1 by A, then take the limit of this expression as A approaches infinity.
When I do all this, I get exactly the final result for H(s) given in your image.