Differential Amplifier's Unusual Behavior
The first thing I noted when I saw your circuit is that the VAS (Voltage Amplification Stage, as you call it) is directly connected to the feedback network and this implies that its gain and its bandwidth are heavily influenced by the value of the equivalent impedance presented by this network: this is the main cause of its "strange" high frequency behavior. There is also another minor slip in the circuit design: I'll show you point by point what is the influences of your design choices on the frequency response of your amplifier, and since I have been somewhat hazy in my former answer, I edited it again, expanding the contents of the last two points


The low frequency zero associated to the feedback resistor \$R_{F2}\$ and the capacitor \$C\$ is too high: calculating the lower cutoff frequency \$f_{low}\$ of the feedback network using the component values you chosen, one gets
$$
f_{low}=\frac{1}{2\pi R_{F2} C}\approx 723\:\mathrm{Hz}
$$
However, considering that the BJT \$Q_2\$ has probably an input impedance which is at most one order of magnitude greater than \$R_{F2}\$, its effect is to push up further \$f_{low}\$ therefore the value of 2 kHz you measures seems completely reasonable to me.
Solution to this problem: rise the value of \$C\$ as much as you can, compatibly with the value of the lower cutoff frequency you want to reach. My advice is to choose \$C\$ in order to get
$$
R_{F2} C=R_b C_{in}
$$
This is the principal problem caused by the interaction (we could say loading) of your VAS by your feedback network. In other words, when you change the value of the feedback by varying the value of \$P_{F1}\$, you change also the open loop gain of the amplifier and thus the gain-bandwidth product \$GBW\$ in a way that it is no more a "constant" characterizing the behavior of the circuit. The core of the problem is the behavior of the gain and bandwith of \$Q_8\$ when \$P_{F1}\$ varies and I'll describe it with the aid of the circuits in following picture:

Part a is the VAS electrical schematic (I used a NPN BJT just because it was easier scannerize and edit :D) of the VAS, part b is the equivalent small-signal circuit, part c is the Miller equivalent circuit: 


I assumed \$C_\mu=C_{bc}+C_c\approx C_c\$,
\$r_\pi=V_T/I_B\approx 1200\Omega\$ where \$V_T=k_BT/q\approx 25mV\$ at \$T=298\:\mathrm{K}\$ (\$25^\circ \mathrm{C}\$), \$k_B\$ is the Boltzmann constant and \$q\$ is the elementary charge,
\$v_2/v\approx -g_mR_{eq}\$ is the voltage gain of the stage,
\$R_{eq}\$ is the equivalent resistance which loads the BJT,
$$
R_{eq}\approx R_c\parallel(P_{F1}+R_{F2})
$$
\$C_\mu(1-v/v_2)\approx C_\mu\$ at least as long as \$v_2/v\geq 10\$.


The Miller effect causes the presence of a pole at the input and one at the output of the VAS, respectively valued 
$$
p_i=r_\pi\left[C_\pi+C_\mu\left(1-\frac{v_2}{v}\right)\right]\qquad p_o= R_{eq}C_\mu\left(1-\frac{v}{v_2}\right)
$$
Changing \$P_{F1}\$ changes the value of \$R_{eq}\$ and in turn the gain \$v_2/v\$ and frequency response of the whole circuit: let's see why.


When your rise the value of \$P_{F1}\$, you get less feedback on the base of \$Q_2\$ but far higher gain from \$Q_8\$ since \$R_{eq}\$ increase. Then \$p_i\$ rises by Miller effect, thus the gain rises but the high frequency cutoff decreases stabilizing the circuit: at the unity gain frequency, the phase response of the circuit is practically identical to the one associated to the low frequency pole introduced by \$p_i\$, even if the circuit is a second order system. 
On the other hand, if you decrease the value of \$P_{F1}\$, you get a higher feedback on the base of \$Q_2\$ but far lower gain from \$Q_8\$. Then both \$p_i\$ and \$p_o\$ get smaller and came closer each other in value: the gain is reduced, but the bandwidth and thus the gain-bandwidth product is highly extended (as you noticed in point 3 of your question, there seem to be no high frequency rollback). This make the circuit behave as a second order system. The scope picture you've taken are illuminating in this respect: 


The first picture on the higher left corner is the step response of you differential amplifier with minimum \$P_{F1}\$: it is the classical response of a second order underdamped system, with heavy ringing and over/undershoot. 
You increase \$P_{F1}\$ and the behavior is still the same but both ringing and over/undershoot start to vanish, as shown in the picture on the higher right corner. 
Continuing the rise of \$P_{F1}\$ you reach the critical value of damping: the rise time of the system is optimal, without ringing and over/undershoot, as shown in the lower left corner picture. 
Finally, the gain rise make the circuit response indistinguishable from the one of a first order system, as shown in the lower right corner picture. 



Solution to this problem: if you do not want to design a feedback network with (almost) constant input impedance in order to avoid heavy loading of \$Q_8\$ (the heart of the VAS), you should use a (possibly complementary) emitter follower buffer and pick the output signal from the emitter(s) of this stage. This frees your VAS from the heavy dependence on the feedback loop impedance and lets you optimize the circuit gain and response.
As explained in the preceding point 2, decreasing the value of \$P_{F1}\$ increases the feedback signal on the base of \$Q_2\$ and decreases as well the gain \$v_2/v\$ of the VAS (referring again to the circuits a), b) and c)): this decrease in gain reduces the Miller effect and improves the bandwidth in such a way that the gain bandwidth product \$GBW\$ increases, so the high frequency cutoff is higher than expected, as you noticed.


Final notes. I agree with the suggestion given by jonk: try to cure as more as possible the "elementary" part of circuit design which involves the biasing of your circuit at the chosen quiescent point. It helps your design to satisfy the requirements you ask it in every condition, even under heavy variation of the semiconductor parameters due temperature and production spreads, since this rules out many trivial causes of erratic behavior. Said that, few years ago I made a discrete devices OP AMP involving a similar stages and it worked very well in the bandwidth from 0 to 10 MHz with a gain of ten or slightly more.