OK. By my estimation, you need to leave shortly (if your haven't already), so here we go.
Yes, F1 is the fuse already in the bike that I'm assuming is dedicated to accessories.
In the schematic, the dots are physical (soldered or otherwise) connections. Whenever a wire crosses another wire, a dot indicates that the two (or three or a million) are connected together. NO dot means they may cross one another, but they are NOT connected.
I fused the 5 VDC regulator circuit to protect it from an overload. It's easier to replace the fuse than to replace the regulator. Some components have built in short protection, in which case, fuses are not necessary. All a matter of component selection and one's willingness to assume that Murphy Law is all a big lie.
The top lead of T1 is the "collector" (if your familiar with the electron tube , or valves as you guys call'em, a collector is similar to a "plate"), The bottom lead (with the arrow) is the "emitter" (or cathode in tube talk [TT]) and the lead at a 90 degree angle jutting to the right is the "base", or "grid" in TT. Basically, the base controls the flow of current through the transistor from emitter to collector (or "holes" from collector to emitter for you solid state theory purists).
T1, with its BASE held at ground by the shunt switch in the jack, (zero potential), will not conduct. Once a plug is in the jack, thereby opening the shunt switch, current can flow through R2 thereby generating a positive voltage (bias) on T1's base, causing T1 to conduct, which in turn causes the LED to light. The small drain (0.1 mA) in the circuit, with R2 shorted, is the current flowing through R1 to ground (with a tiny tiny bit flowing through T1, though not enough to light the LED.
Transistors come in a huge variety. I chose T1 (2N2222) because it is widely available (in fact all the electronic components are available at Radio Shack). It's cheap and reasonably rugged. We can go deeper here if you like.
R4 is, indeed, a current limiting device. It also acts, in conjunction with the inherent resistance of the LED, as part of a voltage dividing network. So, it serves as both a voltage "reducer" and a current "reducer". Marvelous little devices.
I strongly suggest, at this point, that you take a look at an Ohm's Law tutorial. A huge (HUGE!!) element of practically ALL (and I mean
ALL) electronic circuits requires a firm understanding of Ohm's Law.
See
www.electronics-tutorials.ws/dccircuits/dcp_2.html , for example.
R1, with R2, represent a "voltage divider". What that means is together they are taking the 12 VDC and dividing it into two 6 VDC "pieces". One of 6 VDC "pieces" is applied, as a bias, to the base of T1.
How do they do that, you say? They are both 10K ohms (20k ohms in
series). In essence, the 12 VDC is
multiplied by the ratio of the value of R1 divided by the SUM of R1 plus R2 (or 1/2) , thereby
dividing the 12 VDC into 6 + 6 VDCs AND
reducing the current by 1/2.
The juncture of R1 and R2 is only going to ground when there is NO plug in any of the jacks, thus causing the "shunt" switch to be closed, which shorts out R2. This means that the base (conduction controlling element of the transistor) of T1 is being held at zero (or a minus voltage potential, if you like), thereby keeping T1 from conducting (and LED1 from lighting).
Hope I explained all this satisfactorily. AND accurately.
I should add that my wife (of 32 years) often complains that when she asks me for the time, I explain, in great detail, how a watch works...