I have in the past written a short note on higher ramification groups which tries to address this and some other questions. Let me summarize it below. Result numbers refer to the note linked, where I included full proofs. Some credit goes to Johannes Anschutz, whose lectures I followed when learning the subject.
Even if you do not a priori know that there is going to be a nice quotient-compatible numbering, it is still a natural question to ask how the higher ramification groups behave under taking quotients. Specifically, given a normal subgroup $H=Gal(L/M)$ of $G=Gal(L/K)$, what is the image of the ramification subgroup $G_i$ in the quotient $G/H$? Although not immediately obvious, it turns out to itself be a ramification group:
Proposition 1.8. We have $G_iH/H=(G/H)_j$ where
$$j=\frac{1}{e_{L/M}}\sum_{\tau\in H}\min(i_G(\tau),i+1)-1.$$
Here $i_G(\tau)$ is defined so that $\tau\in G_i$ iff $i_G(\tau)\geq i+1$ (this is the classical normalization, it would be a little cleaner without the $+1$, but oh well!), and $e_{L/M}$ is a ramification index. The proof a little computational but not hard, and comes down to understanding how $i_{G/H}$ and $i_G$ are related.
With this proposition at hand, for $\sigma\in G$ let us define
$$\varphi_{L/K}(\sigma)=\frac{1}{e_{L/K}}\sum_{\sigma\in G}\min(i_G(\tau),i+1)-1$$
so that the proposition gives $G_iH/H=(G/H)_{\varphi_{L/M}(i)}$. As should not be surprising from this relation, these functions behave well in towers:
Lemma 2.1. We have $\varphi_{L/K}=\varphi_{M/K}\circ\varphi_{L/M}$.
With this result in mind, it is easy to see that this may give us a more convenient numbering. Indeed, if we define the numbering so that $G_i=G^{\varphi_{L/K}(i)}$ and analogously for $G/H$, we get
$$G^{\varphi_{L/K}(i)}H/H=G_iH/H=(G/H)_{\varphi_{L/M}(i)}=(G/H)^{\varphi_{M/K}(\varphi_{L/M}(i))}=(G/H)^{\varphi_{L/K}(i)}.$$
Thus setting $G^j=G_{\varphi_{L/K}^{-1}(j)}$ gives a numbering compatible with quotients (Theorem 2.3).
Now the only sticking point is, my $\varphi$ is different from your $\phi$! But it turns out that they are, after all, the same value. This comes out of the proof of Lemma 2.1 - $\varphi$ has a very easy derivative, as really you are just counting how many terms in that sum are increasing, and the answer is precisely $|G_i|$. The integral formula then merely comes down to integrating this derivative, equal to $\frac{|G_i|}{e_{L/K}}=\frac{|G_i|}{|G_0|}=\frac{1}{[G_0:G_i]}$.
