Determine I through load resistor in circuit with diodes in series
If you rearrange your formula, you get
$$
(U_m - 2U_T) = I_m (2R_d + R)
$$
which looks like
$$
V = IR
$$
where \$R\$ is the sum of all of the resistors in the loop and \$V\$ is the voltage across that summed resistor.

Kirchoff tells you that the voltage across this resistor is the same as the voltage looking at the rest of the circuit, which is \$U_m - 2U_T\$, because there must be a voltage drop across each of the diodes for current to flow.