This circuit doesn't protect anything. The transistor only adds to the stress on the power supply by turning on, and providing one more path for current to flow.
If the power supply is an ideal voltage source, then the transistor does not reduce the current which flows to the shorted output. The only way this circuit reduces the current is by adding to the load, so that the power supply's voltage sags more.
By itself it doesn't regulate in any way.
However, suppose that the input side is driven by a regulated power supply which already features overcurrent protection. Then a circuit similar to this one can be used as a "crowbar" to deliberately trip the mechanism long before the device draws that much current. Say the power supply shuts down at 1A (or limits current to 1A), but you want that limiting or shutdown to happen when the load device draws only 100 mA. Then you can pick the sensing resistor so that the transistor fully turns on at 100 mA (around 7 ohms). Do not have any resistance in the collector circuit, so that the transistor shorts the power supply.
There is a better ways to use a PNP transistor to limit current to 100 mA. Rather than shorting out a power supply to trigger its limiting/shutdown, we can simply build a current limiter:
Here, the diodes are forward biased, establishing about a 1.4V drop. Diodes are used because their voltage will not vary significantly with a varying IN voltage, providing a stable reference. The transistor's BE junction needs 0.7V to turn on, leaving the remaining 0.7V across R1. We can choose the emitter resistor R1 to limit the OUT current to 100 mA using \$R = V/I = 0.7/0.1 = 7\Omega\$.
The value of R2 is much larger: it has to only conduct enough for the transistor to have enough base current drive and let some current through the diodes.
In this circuit, if OUT is short circuited to ground, still only 100 mA will flow (for the 7 ohm value of R1). In that case, the transistor will have some voltage across it, and will be dissipating heat. A power transistor with a heat sink may be needed. For instance, suppose the \$V_{CE}\$ voltage is 10V when there is a short circuit. In that case, the transistor is dissipating 1W.
You can use a single Zener diode instead of the two silicon diodes. R1 has to be adjusted then to keep the same current, in light of the Zener voltage.
So you have a bi-color LED for which both colors are known to work when connected to batteries. You've asked for a way to connect them to a motor such that the colors indicate the rotational direction.
What you haven't specified is how you are powering the motor (same two AA batteries?). You didn't specify how you are reversing the direction of the motor (is it a mechanical switch or transistors?). That's the purpose of a schematic: to illustrate how everything is connected.
You should definitely invest some time and effort into learning how to draw a schematic. It doesn't have to be fancy, and it doesn't even have to be drawn with software. You could draw on paper and take a picture. Its whole purpose is to communicate to other electronics engineers what you have in your circuit. Have you heard the phrase "A picture is worth a thousand words"? It's never been more true than in electronics as far as schematics are concerned.
So let's start with an LED. This is a light-emitting diode, so it uses the schematic symbol for diode plus two arrows to indicate light.
Your 2-color (bipolar) LED is two of these with either the anodes or cathodes tied together (this diagram is common anode):
I'll assume that your motor is designed for 3 volts. Even though you can light the LEDs with this voltage as well, you should use current limiting resistors to prevent the LEDs from burning out. Without knowing specifically what LEDs you are using, we will assume that each color \$V_f\$ of 2.2V, a common value for red/green bipolar LEDs. We will also assume that they have a standard \$I_f\$ of 20mA. If you're not familiar with driving LEDs (forward voltage and forward current) you should look at some of the other questions here that address how to power LEDs properly.
With these values, we can select the value of current limiting resistor using Ohm's Law: \$R = E / I\$.
$$R = \frac{3 - 2.2}{0.020} = \frac{0.8}{0.020} = 40\Omega$$
If you only plan on having one color lit at a time, you can use one resistor on the common pin:
You can use a DPDT (double pole, double throw) switch to change the direction on the motor with no other components:
However, you can't add a bipolar LED easily because it doesn't change color based on polarity, but rather on which cathode (or anode) is connected rather than open.
You could add two separate LEDs in reverse bias to each other easily like this:
To add the bipolar LED, you'll need a 3PDT switch so you can use the third pole to alternate the cathode (or anode) of the LED:
Now, none of this may actually be what you are using or how you intend to do it, but hopefully it gives you something to work from, and shows you the importance of diagrams and schematics. Good luck in your project!
Edit:
As @Wouter pointed out in the comments, you can do this with a DPDT switch:
Best Answer
That's not fungal growth; it's corrosion of the solder joints and flux residue around them. You can clean it off with normal cleaners like isopropanol.