Does the phrase "instantaneous frequency" make sense?

Yes, you can make sense of the concept of instantaneous frequency. Imagine a time-dependent phase $\mathrm e^{i\phi(t)}$. For $\phi=\omega t$, we would call $\omega$ the angular frequency, and this is $\mathrm d\phi/\mathrm dt$. Thus we can regard $\mathrm d\phi/\mathrm dt$, the rate of change of the phase angle with time, as an instantaneous frequency. We can still do this if the amplitude is also time-dependent, by ignoring the amplitude and focussing on the rate of change of the phase.

If we only have a real signal, we would in general have to make an arbitrary decision what part of its change is due to a change in "amplitude" and what part is due to a change in "phase", since taking the real part of different complex signals can lead to the same real signal. However, there may be situations in which it makes sense to regard the signal as having a constant amplitude, and to regard all changes as changes in phase. The example you give is such a case, since the signal oscillates between $+1$ and $-1$. In such a case, we can consider the signal as the real part of a complex signal of constant amplitude.

That is, given a real signal $f(t)$ and a constant amplitude $A$, we can consider the complex signal $f(t)\pm\mathrm i\sqrt{A^2-f(t)^2}$, where the sign changes whenever the signal reaches an extremum with $f(t)=\pm A$. Then we can calculate the phase as $\phi(t)=\arccos(f(t)/A)$, where we let the phase wrap around and increase beyond the usual range $[0,2\pi]$ to avoid discontinuities. Thus we can define an instantaneous angular frequency as $\omega(t)=\mathrm d\phi(t)/\mathrm dt=\mathrm d/\mathrm dt(\arccos(f(t)/A)$. In your case, the result would be $\omega(t)=\mathrm d/\mathrm dt(t^2)=2t$.


To add a slightly different viewpoint to @Joriki's excellent answer, communications engineers often restrict the notion of instantaneous frequency to (real-valued) signals that can be expressed as $\text{Re}\{a(t)\exp(j\phi(t))\}$ where $a(t)$ is a (possibly complex-valued) continuous signal that is varying much more slowly with $t$ than $\phi(t)$ is, and $\phi(t)$ is a continuous function of $t$ that is differentiable for all $t$ except possibly for a discrete set of time instants. Then the instantaneous (radian or angular) frequency is the derivative of the phase. For the OP's problem, we have $a(t) = -j, \phi(t) = t^2$ and everyone is in agreement that the instantaneous frequency of $\sin(t^2)$ is $2t$.