Barkhausen criteria: which is the overall effect if only the magnitude criterion is met?
I think, I can agree to everything you have explained. That means: Yes, the Barkhausen condition may be applied to steady-state oscillations only. With other words - if the amplitude condition is met (|T(jωo)|=1 or even >1) and the phase condition is NOT met, we can have decaying or even rising oscillations immediately after power switch-on. But - at which frequency? In reality we have the following situation: 

The system will always try to oscillate at a frequency where the phase condition is met. This is the only possibility to allow self-excitement of a system with positive feedback. And this frequency very often will be NOT exactly the frequency (wo) where the amplitude condition is fulfilled per design (or "over-fulfilled"). Hence, the system will try to oscillate at another frequency (let`s say w1). If the amplitude condition is still fulfilled at this new frequency w1 - we will have rising oscillation amplitudes (until any limiting mechanism will come into play). However, if the loop gain magnitude will be &lt;1 at the frequency w1 we will observe damped oscillations (or even no oscillations).

Answer to your second question: If Barkhausens condition (amplitude and phase) is fulfilled for one single frequency only, there is no chance that oscillations at other frequencies are build up.    

EDIT 1(example)

Here is a very illustrative example which exactly meets the subject of your question:

There is a harmonic oscillator consisting of  two active opamp integrators in a closed loop (one is inverting and the other one is non-inverting). Hence, under ideal conditions we have a loop phase of 0 deg over a rather broad frequency range. However, for real opamps the loop phase will cross the 0 deg line at one single frequency fo only. We will observe that  the loop gain magnitude will cross the 0 dB line at another frequency f1 which is always larger than fo (f1>fo). This is due to the opamps finite open-loop gain and corresponding phase shift.

That means: The loop gain magnitude at the frequency fo will be > 0 dB and the Barkhausen condition is (over-)fulfilled. The circuit oscillates at a frequency fo (where the phase condition is fulfilled). Note that for all frequencies below fo the loop gain is even larger than at f=fo. Nevertheless, the frequency for steady-state oscillations is f=fo.

Final remark: In particular, this oscillator topology is most suitable for explaining why and at which frequency a linear oscillator will oscillate. 

EDIT 2 (Comment to the example cicrcuit):

Perhaps the following considerations may further help to answer your (very intelligent) question. If you try to use one of the available simulation programs you will find that the two-integrator oscillator will NOT continuously oscillate in case of simplified first-order models for the opamps. Why not? Because the loop phase will meet the 0 deg line for infinite frequencies only. For all finite frequencies, the loop phase is very close to 0 deg but not exactly zero (second-order system only). For this reason, after power switch-on we will have slowly decaying oscillations. 

To allow continuous oscillations, a simple resistor across the feedback capacitor will turn one of the integrators into a two-pole system - and the loop gain now is of 3rd order with zero loop phase at a finite frequency.