Secondary/Failover power for board
Here is a solution with discrete components:

NOTE: The component selection did not consider your actual current and power values. The MOSFETs and the diode would have to be rated accordingly

Essentially how it works:

The primary power supply takes precedence over the secondary one, by controlling the latter via a PMOS based ORing, which is connected to two inverters. 
If the primary's voltage is high enough, it will turn on \$Q_2\$, which turns off \$Q_1\$, which in turn turns off the back-to-back MOSFETs \$M_1\$ and \$M_2\$. The threshold, at which you want to have the transition from one power supply to another, can be determined by the resistors \$R_4\$ and \$R_5\$ and the BJT's \$V_{BE}\$. For a more precise transition point, let's say, if the primary supply voltage falls below \$4 V\$ the secondary power supply is connected, you can use a comparator.

As you can see in  the waveform, the load is only sourced with the current from the active power supply.

EDIT #1


Reduce the turn-on / off time of the ORing by charging its gate capacitance to no more than \$4.7V\$ (No need for overcharging it). It also reduces power losses during the transition.
Asymmetrical charging and discharging of the gate capacitance via \$R_1\$ and \$R_6\$ adds hysteresis to the transition e.g. \$V_{OFF}=6.4V\$ and \$V_{ON}=8.13V\$
Resistive load replaced with a constant \$10W\$ power load