Analyzing a transistor relay driver
It's not clear from your diagram that the input is supposed to be \$+12\:\textrm{V}\$ in order to activate the relay. But let's assume you are right about that detail. (I don't have any reason to disagree with you about that.)



Let's start with the \$2.2\:\textrm{k}\Omega\$ resistor from base to ground. This resistor helps to hold the base near ground (BJT = OFF) unless there is a definite activation signal at the input; helps to protect the BJT base from possibly damaging reverse voltages; and provides a simple DC path to ground from the BJT base (which otherwise may only have a diode path to ground.) Imagine a case where somewhere in the circuit there is also a source of \$-12\:\textrm{V}\$, also. And suppose this accidentally were to be connected to the activating input (left side of the \$4.7\:\textrm{k}\Omega\$ resistor) by a ham-handed idiot. The BJT base-emitter junction would be reverse biased and therefore directly exposed to the full voltage. But most small signal BJTs cannot tolerate more than \$5-6\:\textrm{V}\$ of reverse voltage before avalanching (and possibly resulting in permanent damage.) In this case, the \$2.2\:\textrm{k}\Omega\$ resistor helps a lot by keeping the reverse voltage low enough to be safe. I could go on. But the idea is to help make the input robust.

In addition to the protection/robustness, their choice arranges things so that 32% of the applied input voltage becomes the Thevenin applied voltage, applied through an equivalent series resistor of \$1.5\:\textrm{k}\Omega\$. So the base current will be \$I_B=\frac{0.32\cdot V_{IN}-V_{BE}}{1.5\:\textrm{k}\Omega}\$. In your example, this is about \$2\:\textrm{mA}\$ of base current for an applied \$+12\:\textrm{V}\$ activation voltage.

Your relay, as you describe it, wants \$\frac{12\:\textrm{V}}{392\:\Omega}\approx 31\:\textrm{mA}\$. Let's call that \$30\:\textrm{mA}\$. With \$2\:\textrm{mA}\$ base current, this means \$\beta=15\$. That's often fine for a saturation \$\beta\$ using small signal BJTs. So I'd be happy to leave that base drive arrangement alone.

But in reading your "analysis," you seem to imagine that anything better than about \$100\:\mu\textrm{A}\$ base current would be enough. You argue this on the basis of \$\beta=300\$ for the 2N3904. The problems here are: (1) that while \$\beta\$ for any one BJT might be relatively flat over a wide dynamic range of collector currents, the \$\beta\$ of any particular BJT will differ from each other by as much as a factor of 2 or so; and (2) these higher "flat" \$\beta\$ values only apply while the collector-emitter voltage is more than a volt or two and you want to use it as a switch, so this \$\beta\$ value no longer is applicable; and (3) so therefore you need to plan on using a guaranteed saturation \$\beta\$, which will be a lot less. I'd consider a \$\beta\le 20\$ for the 2N3904 as "good enough." Your circuit's value of \$\beta=15\$ is nicely in that pocket. So I'd keep it.



Your relay not only includes the resistance you mentioned but also an inductance. In the very first moments of activation, before the resistance can dominate things, the inductance will limit the rate of change in the current. In short, it will take a little bit of time for the current to reach the limiting value of about \$30\:\textrm{mA}\$ discussed earlier because the inductance has "inertia" (relative to rapid current change) to it and keeps the current from instantly reaching that value. This inertia works both ways, though. So when you try and remove the current supply by switching the BJT off, the inductance of the relay resists this just as vigorously.

The sign of the voltage across the inductance of your relay is based ONLY on the direction of the change in current. So when turning it on, the applied voltage is of course positive as the current "ramps upward." But when turning it off, the sign of the voltage reverses. Now, the magnitude of that reversed voltage is a matter of just how fast the current declines. If it instantly declines to zero (impossible), then the reverse voltage would be infinite. We both know that doesn't happen. But this is why some kind of protection helps here. You need to provide some "soft landing" method so that the inductance doesn't induce a very high reverse voltage that damages whatever is near to it. A diode does this very well. Some people will use a zener plus a diode, to provide a higher voltage and therefore a faster decline in the current.

That covers the diode across the relay. The diode across the collector-emitter junction of the BJT is a different story and probably isn't necessary. But all BJTs can also be placed in reverse-active mode and this diode would keep the BJT's collector from being driven sufficiently negative that it could be turned on through the \$2.2\:\textrm{k}\Omega\$, if for some reason the collector were driven more negative than ground. I'm guessing here that perhaps someone might add it (if not by you to just ask a question) because the collector node is brought out to a screw terminal and... once again... the designer is protecting against "idiots" who might cause a problem when wiring things up (depending on a fuse somewhere perhaps to blow in order to "fix" the problem created by the idiot.) But I'm just a hobbyist and perhaps a professional engineer here will teach me a thing or two on this aspect.



I'm not so interested in writing a book on oscilloscopes. And there are much better people here to write one, if that were needed. But I think you will be fine just learning to use your scope on this circuit. You can use the 10X probe, if you have it (you should.) At these early stages you will have the most problem starting out and the greatest need to learn about triggering, I think. Once you are comfortable with that, the rest will follow more easily. But there are lots of tutorials around.