This is a nice question and has a rather illuminating answer.
Let's start simply, if I may, by considering 1 dimension rather than 3 and an infinite square well potential rather than the $1/r$ Coulomb field. If the well goes from $-a/2$ to $+a/2$ then it has solutions $\psi_n(x)=\sqrt{2 \over a} \cos{n \pi x\over a}$ when $n$ is an odd positive integer and $\psi_n(x)=\sqrt{2 \over a} \sin{n \pi x\over a}$ when $n$ is an even positive integer, and these have energies $E_n={\hbar^2 n^2 \pi^2 \over 2 m a^2}$ and the full wavefunctions including the time dependence are $\Psi_n(x,t)=\psi_n(x)e^{-iE_n t /\hbar}$. Standard stuff.
Notice - it's trivial but important - that the mean position of the electron $\langle x \rangle =\int {\Psi_n}^*(x,t) x \Psi_n(x,t) dx$ is zero for all $n$, as $x$ is odd and $\psi(x)^2$ is even .
Now consider an electron which starts in an excited state, say state 2, and decays to a lower state, say state 1, the ground state. Initially it is in $\Psi_2(x,t)$ and finally in $\Psi_1(x,t)$. In the middle it is in some superposition of the two. $\Psi(x,t)=A\Psi_1(x,t)+B\Psi_2(x,t)$. The energy during this (short) intermediate period is not defined but that's OK because of the uncertainty principle. $A$ and $B$ are functions of time and normalised to one, but these details don't concern us right now.
Now $\langle x\rangle$ is $\int \Psi^*(x,t) x \Psi(x,t) dx$ which is $\int (A{\Psi_1}^*(x,t)+B{\Psi_2}^*(x,t)) x (A{\Psi_1}(x,t)+B{\Psi_2}(x,t)) dx$
This contains terms $|A^2||\psi_1^2|x$ and $|B^2||\psi_2^2|x$ which vanish as before, but it also contains a couple of cross terms
$[AB {\Psi_1}^*(x,t){\Psi_2}(x,t)+ AB {\Psi_1}(x,t){\Psi_2}^*(x,t)]x$
Putting in the expressions for $\Psi_1$ and $\Psi_2$ turns this into
${2 AB\over a}[e^{i(E_2-E_1)t/\hbar} + e^{-i(E_2-E_1)t/\hbar}]x \cos{\pi x \over a} \sin{2 \pi x \over a}$
The space integral does not vanish, as it is the product of an even function and two odd functions. The time dependence looks like $\cos(E_2-E_1)t/\hbar$.
Now remember that the electron has a charge. What the maths tells us is that during the transition there is a dipole moment which oscillates with frequency $\omega = (E_2-E_1)/\hbar$, i.e.$f = (E_2-E_1)/h$. It is behaving as a little dipole radiator oscillating at just the right frequency to emit the EM radiation corresponding to the energy transition (i.e. the photon)
Notice that we pick up for free the selection rule that such transitions can only occur between odd and even states. In 3D such rules are more complicated but are basically just saying that the integral of the product of the space parts of the two states involved must not vanish.