You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
Switching an inductor may make it work like a boost converter, and thus temporarily generate a higher voltage.
One way of making sure that the input voltage to the LM338 converter is no higher than allowed, is to wire a high-wattage (between 600W and 5 kW) TVS diode with an appropriate clamping max voltage biased against the input voltage and ground.
Also, if you're switching on the measured/actual output voltage, then you can get into oscillation. It would be safer to switch on "desired" output voltage, and let the regulator make sure it gets there. You could perhaps feed the bias voltage from the regulator ADJ into your microcontroller with a resistive divider to achieve this. Or you could just use a comparator with a pre-set trip point to drive/un-drive the relay.
Best Answer
With a synchronous SMPS, you must add your own "OR-ing diodes" from the SMPS output to the power bus. It doesn't matter that most SMPS have diodes on the input. Most synchronous SMPS, when it sees its output voltage is "too high", will pull current from the output and store it on the "input" capacitor. If a person pulls the output voltage too high for too long, that "input" capacitor will eventually fail from over-voltage.
If a power supply -- such as practically all alternators -- have diodes on the output, then those diodes can do double-duty as the "OR-ing diodes".
With "isolation diodes", sometimes called "OR-ing diodes" (either the internal built-in ones of an alternator, or the external ones you'll probably need to add to a switching regulator, or both), you can connect two power supplies to a system and it will work just as you expect -- whichever power supply is turned on and generating the higher voltage will supply almost all the power to the load.
You may want smoothing capacitors at the load. When the SMPS is turned off, no current flows from the power bus "backwards" through the output OR-ing diodes into the smoothing capacitors of the SMPS, so those capacitors are effectively isolated and do nothing to smooth the voltage produced by the alternator.
"Can power supplies be connected in parallel for redundant operation?"
"Considerations for parallel power supply outputs"
"Using Isolation Diodes for Parallel-Redundant Operation"
"Connecting Switching power supplies in parallel is possible if certain precautions are observed."