Confusion regarding signal amplifier
Preface
In the following, I'll respond in kind to your question. That means I won't bother criticizing the specific values you used or talking about better (in some ways, at least) topologies. I'll just take them as given. As I wrote in comments, I'm actually pretty impressed at how well-formed your question is. Perhaps it's also an omen about just how smart you are, too, even if you can't lay claim to an intention here. You've focused on the question you have in mind and your schematic is the least number of parts required to get the basic idea across. Nice.
Warm-Up
Engineers bandy around all kinds of terms that are meaningful to them. CC, CE, CB are mnemonics for arrangement classes of the BJT stage, whatever it may be. You'll read those all over the place. But of course without studying them for a while to see what it covers and what it does not cover, it's just mumble-words. Worse, other terms are also used, such as cascode (CB), emitter-follower (CC), and the like.
The bottom line is that with only three pins a single-BJT amplifier stage pretty much has the requirement of a single-ended input signal that shares one of the BJT pins in common with the output signal. (Single-ended is another one of those phrases that simply means that one of the signal wires [there must always be two] is shared in common with an output signal in some, not necessarily obvious, fashion.)
Signal Voltages
All voltages are relative. Don't ever forget that. So when you talk about a signal voltage, it always has two points where a voltmeter or an oscilloscope would need to be applied in order to see well. If you look at your left-hand side schematic, you should be able to see that the signal is applied across the base and emitter and that the output would be taken from the collector and emitter. (The emitter is &quot;at ground&quot; so there is a hidden implication here.) This means that the emitter is &quot;in common&quot; and therefore this is a CE stage -- a common emitter. All of us know that when you write CE stage that it should call to mind qualities about CE stages that they share and/or don't possess.
Another thing to always keep in mind is that BJTs are just like MOSFETs in the sense that they are voltage controlled current sources/sinks. They are unlike MOSFETs in that their control voltage incurs a required base-emitter junction recombination current that is in direct proportion to the applied control signal voltage to keep things working right; where the MOSFET only requires a gate current when the applied control signal voltage is changing. (Also, MOSFETs are squared-law devices, while BJTs are exponential [due to something called the Shockley equation that I'll be talking about, shortly.])
Left-Side Stage
Where you are getting lost in thinking about the left-hand schematic is when you imagine that you can analyze it thinking about the base and emitter currents. You can't. It's just the wrong way to look at it.
What's happening is that your signal (which luckily has to go through a capacitor, so that the BJT stage can naturally find a DC operating point -- good job there) is varying the voltage at the BJT base-emitter junction (the input signal pair.) This varying voltage is causing a varying collector current. And, as it turns out, you have to look at the Shockley diode equation. (It originated with diodes but it turns out that a very similar equation applies to BJTs in their active mode [sorry, yet another term to learn well.])
You can refer to this answer I gave earlier for more details about how to derive the dynamic emitter resistance from the Shockley equation, but it turns out there actually is (with respect to variations of the base-emitter voltage with respect to variations of the collector current) a dynamic emitter resistance that is often important when thinking about a BJT amplifier stage. (Also see this site for more discussion about this new term.) This is given the symbol: \$r_e^{'}\$.
Again, for slightly more on the voltage gain expressions, you can go to this answer of mine. But for your left-hand schematic the relevant voltage gain (your circuit is unloaded at the collector, so it applies well) is:
$$A_{v_{_\text{UNLOADED}}} =-\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{V_T}=-\frac{R_\text{C}}{r_e^{'}}$$
In using LTspice with a 2N2222 BJT in your first schematic, the DC operating point comes out to \$V_\text{C}\approx 4\:\text{V}\$. (It uses \$27\:^\circ\text{C}\$ as the operating temperature, so \$V_T\approx 25.86492\:\text{mV}\$.) This works out to \$A_v\approx -300\$. For validating this rough computation, I used a capacitor value in LTspice of \$470\:\mu\text{F}\$ to avoid its reactance being a complication -- your value has a reactance at \$1\:\text{kHz}\$ of about \$180\:\Omega\$. The results give me \$A_v=-213\$.
Now LTspice also tells me that \$r_e^{'}\approx 1\:\Omega\$. But when I look at the BJT model for the 2N2222 that LTspice is using I also find that there is a small emitter bulk resistance (perhaps confusingly called \$r_e\$) that is \$200\:\text{m}\Omega\$. Note that this is already a 20% error, or so, and adding that value to \$r_e^{'}\$ would reduce the voltage gain to \$A_v=-250\$. And there are still other parasitics and model details that LTspice uses in the BJT model that further decrease the voltage gain magnitude, just a little bit more.
So, this is how you should perceive the voltage gain for this kind of stage.
Right-Side Stage
In this case, it's a lot easier to answer your question. Whenever the base voltage moves upward, the emitter also moves upward with it. Same thing in the other direction. That's because there is a base-emitter forward voltage present that is &quot;operating&quot; the BJT. In this case, LTspice says it is about \$730\:\text{mV}\$. (But that will vary with different BJT models, a little bit.) This voltage is mostly unchanging. So the difference between the base and the emitter remains &quot;fixed.&quot;
If that were entirely true then it would follow that whatever AC signal you applied to the base would appear, unchanged, at the emitter. Then \$A_v=1\$. But it turns out that once again the Shockley equation comes in to complicate it a little. As the applied signal pulls up on the base, thus also pulling up almost equally (the almost part is what I'm talking about now) on the emitter. When the emitter rises, so also does the voltage across that emitter resistor, \$R_3\$. That means more emitter current, which means more collector current, too. And the Shockley equation says that with a higher current then the base-emitter voltage also has to be larger. But that means that the emitter didn't completely follow the base, upward, but because of the slight increase in the voltage difference, only came up almost as much. So the actual voltage gain in this configuration is \$A_v\lt 1\$. But not by much. Here, LTspice reports \$A_v\approx 0.98\$.
Oh, and the right-side schematic is a CC stage. You can &quot;see this&quot; since the input signal is applied to the base and something and the output is taken from the emitter and something. Now, the something has to be the other pin that we didn't mention, the collector. That's the only choice.
But you can also just imagine it in your head, too. Your circuit would do exactly the same thing if, instead of attaching one lead of your input signal to ground you instead tied that same end up to the power supply rail of \$12\:\text{V}\$. Then you can easily see that the input signal can just as well be said as applied across the base and collector. Similarly, the output could also be considered as taken between the emitter and collector. So the collector really is in common here.
Here, this might be an example of that &quot;not necessarily obvious fashion&quot; I mentioned, earlier.
This circuit has the emitter following the base. So it is also called an emitter-follower. The choice between CC and emitter-follower will be about emphasizing something the author wants to pull forward in the conversation. If the most important thing is that the emitter follows the input signal, then emitter-follower is used. If, perhaps, the fact that the input resistance is \$\beta+1\$ times the emitter resistance, then perhaps CC might be chosen. It's based on experience and what are the central ideas needed to quickly grasp the main point of using the stage.
Summary
Note that in all these cases we are comparing the AC output voltage with the AC input voltage. But that isn't always why these different stages are used.
For example, your right-hand schematic is often used to increase the current-compliance of a prior BJT stage. That's because only a little current from the prior stage is required to drive this stage's base, and its emitter is almost an exact replica of the input, but here the emitter is capable of supplying so much more current (getting most of that from the collector which is directly tied to the power supply!)
So it's not all about \$A_v\$, either.