The diagram can be rearranged so that \$H_2G_2\$ is an inner loop and $$\frac{G_1H_1G_2}{1+G_2H_2}$$ is the outer loop.
It is required for the outer loop to be stable. The inner loop may be unstable with the full loop being stabilised by controller \$G_1\$. This is generally only done when absolutely necessary to stabilise an unstable plant.
Practically speaking, both loops should be required to be stable in a system such as this with standard op-amps.
As a rule of thumb for design simplicity, the \$G_2H_2\$ loop is designed to be stable with significantly higher bandwidth than the final loop bandwidth so that its phase delay can be neglected in the design of the outer loop; i.e. its phase/magnitude contribution to the overall loop is negligible at crossover.
To the OP's Update: The point of which loop to analyse: A straightforward analysis of the OP's loop (also the Updated equivalent arrangement) shows the stability analysis to be:
$$G_2(G_1H_1+H_2)+1=0$$
for the rearranged circuit with the inner loop we get:
$$\frac{G_1G_2H_1}{1+G_2H_2}+1=0$$
Applying simple algebra we see that it's the same result, as expected:
$$G_2(G_1H_1+H_2)+1=0$$
Therefore you can analyse the loop several ways and get the same result. Considering the loop to consist of an inner and outer loop, is however, very convenient.
UPDATE:
Here's the analytic justification for the loop stability equation to be $$G_2(G_1H_1+H_2)+1=0$$ Apologies for quality. I hope to enter as native format when I get time:
When evaluating the stability of a control system, it's most insightful to plot the loop gain, either as a Bode plot or as a Nyquist plot. They do not differ in this respect.
The Nyquist plot is useful for employing the Nyquist stability criterion. In summary, loop gain encirclements of the point \$(-1, 0)\$ on the Nyquist plot indicate instability. Unless the system was already unstable (i.e., has RHP poles), in which case a counter-clockwise encirclement must be made for each RHP pole.
On a Bode plot, the usual technique for evaluating stability is to investigate the gain margin and phase margin of the loop gain. If both of these values are greater than zero, then the system is stable (as long as it doesn't have RHP poles). This technique isn't quite as general as the Nyquist criterion, but for the vast majority of control systems it's good enough. It's possible to evaluate the Nyquist criterion by looking at a Bode plot, but it's more difficult.
So, why would you evaluate stability with a Bode plot when the Nyquist criterion is more general? Because the Bode plot gives you a lot of insight that the Nyquist plot doesn't. The Bode plot shows gain and phase versus frequency, helping you identify what frequencies to place compensating poles and zeroes, as well as lending insights into closed-loop response that are impossible to see on a Nyquist plot (such as closed-loop bandwidth).
Finally, once you've determined your system is stable, you can re-use the Bode plot for a meaningful demonstration of closed-loop response as well. Plotting closed-loop gain on a Nyquist plot isn't nearly as meaningful.
Best Answer
The procedure is quite simple.
When you have the BODE plot for the LOOP GAIN (open-loop transfer function) you only must shift the magnitude plot up or down until the desired phase margin is met. This shifting will not influence the phase plot. As a result of this shifting the frequency where the magnitude crosses the horizontal axis (0 dB) will be shifted as well.
Of course, the amount of shifting gives you the required constant gain (in dB).
Remember that the loop gain is nothing else than the product of the remaining blocks (better; the corresponding transfer function). In the BODE plot, both functions are to be added, of course (due to the log display).
Are you familiar with the definition of the phase margin? It is defined NOT for the closed-loop but for the loop gain!