How to design a good common-collector amplifier?
The problem is that the NPN transistor can only &quot;source&quot; the current and cannot &quot;sink&quot; current (current can only flow out of the emitter and never flow into emitter).
In your circuit, \$R_E\$ resistor can only &quot;sink&quot; current.
Thus, the positive swing current is &quot;provided&quot; by BJT. But for the negative swing, we have a problem because now \$R_E\$ resistor needs to be able to &quot;sink&quot; this negative current.
What this schematic is trying to show in a simplified manner:

Negative clipping occurs when \$I_E \$ current reches \$I_E = 0A\$ (transistor is cutt-off).
Thus, the maxumum negative voltage you can get at the outpus is equal to:
\$V_{OUT_{MAX}} = V_{C_2} \times \frac{R_L}{R_E + R_L} = I_{EQ} \times R_L||R_E\$
Where:
\$V_{C_2}\$ is a DC voltage across output capacitor (DC voltage at the Emitter)
\$I_{EQ}\$ - Emitter quiescent current.
Why does the emitter follower clip?

How can I design a CC amplifier circuit that has a better output
swing?

You can reduce the \$R_E\$ resistor value or use a push-pull emitter follower.
How does this Push-Pull amplifier work?

How does it get this &quot;weird&quot; shape?

In your circuit, the DC voltage at the base will be around \$V_B \approx 7V\$.
Thus, for the positive swing, the voltage at the base will be:
\$V_{B_{DC}} + V_{IN} \approx 7V + 4V \approx 11V\$ at the peak of the input signal amplitude.
And this means the transistor will enter the saturation region because of \$ V_B &gt; V_C +  0.5V\$. And in saturation, the base-collector is in forward conduction.
And this is why you see this &quot;weird&quot; shape.