According to Peskin and Scrhroeder the pion decay constant $f_\pi$ is defined via the following matrix element $$\left\langle0|j^{\mu5a}(x)|\pi^b(q)\right\rangle=-if_\pi \delta^{ab} q^\mu e^{-iqx}$$ while $j^{\mu5a}=\bar{Q}\gamma^\mu\gamma^5\tau^a Q$ with $Q$ the up-down quark doublet $Q=(u, d)^T$. The point is, that intuitively the definition seems to imply that $f_\pi$ is a kind of amplitude for a pion to behave like a pair of quarks and vice-versa.
This intuition works, for example, in the leptonic decay of a charged pion, where the amplitude $\mathcal{M}(\pi^+\to l^+\nu_l)\propto f_\pi G_F$. The first multiplier I interpret here as the amplitude for the pion, a bound state of quarks, to decouple into two free quarks, and the Fermi constant $G_F$ then describes how those two quarks decay into a lepton pair.
My question is the following. How can I understand the appearance of the negative powers of $f_\pi$ in some other hadronic amplitudes? For example, take the $\pi^0\to2\gamma$ decay (which bothers me most, practically) with the amplitude $$\mathcal{M}(\pi^0\to2\gamma)\propto \frac{e^2}{f_\pi}$$
One can visualize this process via the Feynman diagram where the pion first splits into a quark pair, then one of the quarks emit photon and, after, annihilate with the other quark emitting the second photon. The factor $e^2$ is natural since there were two photons emitted. But I would expect the $f_\pi$ factor to appear in the numerator, just as in the case with a leptonic decay. What is the reason here that it is in fact the negative power of $f_\pi$ in the correct formula?
Answer
How to compute pion interactions in general? Since pions are (pseudo)goldstone bosons, the procedure is following: starting from lagrangian which contains quark fields $q(x)$ (and suppose that you've integrated out gluon sector), you need to extract pion degrees of freedom from them, namely $$ q(x) \equiv U\tilde{q}(x), $$ where $$ U = e^{i\gamma_{5}\frac{\pi_{a}t_{a}}{f_{\pi}}} $$ There you just need to subtitute nonzero VEV $\langle |\bar{\tilde{q}}\tilde{q} |\rangle$.
From the described picture follows that the basic object of pion effective field theory is $U$, which depends on argument $\frac{\pi}{f_{\pi}}$.
In general the relevant effective lagrangian for free pions which is needed to answer your question has the form $$ \tag 1 S = \int d^{4}x\left(\frac{f_{\pi}^{2}}{4}\text{Tr}\left[ \partial_{\mu}U\partial^{\mu}U^{+}\right] - f_{\pi}^{2}m_{\pi}^{2}\text{Tr}\left[ U^{\dagger} + U - 2\right] + ...\right) + N_{c}\Gamma_{WZ}, $$ Here $$ \Gamma_{WZ} \equiv \frac{i}{240 \pi^{2}}\int d^{5}x\epsilon^{ijklm}\text{Tr}\left[U\partial_{i}U^{-1}U\partial_{j}U^{-1}U\partial_{k}U^{-1}U\partial_{l}U^{-1}U\partial_{m}U^{-1}\right] $$ is the Wess-Zumino term, which captures all the information about anomalies of underlying theory, and dots represent higher derivative terms and mixed terms.
Note the presence of $f_{\pi}^{2}$ at the first term in $(1)$ (it is needed for canonical form of pions kinetic term) and an absense of such quantity in the Wess-Zumino term (since in fact it is just the number).
The short formal answer on your question is following. In order to describe the mentioned processes - $\pi^{+} \to l^{+} \nu_{l}, \pi^{0} \to ll^{+}, \pi^{0} \to 2\gamma$ - we need to elongate the derivatives in $(1)$. The in turns out that the first two processes are mediated by the kinetic term in chiral effective field theory action $(1)$, which has $f_{\pi}^{2}$ factor in the front, while the general contribution into process $\pi \to \gamma\gamma$ is made by the Wess-Zumino term, which hasn't any dimensional factors in front of it. The amplitudes for the first two processes are proportional to $f_{\pi}$, while for the latter it is proportional to $f_{\pi}^{-1}$.
My statement that after gauging and adding the lepton part such action contains information about processes $\pi \to \mu \bar{\nu}_{\mu}$ and $\pi \to \gamma \gamma$.
When gauging the first terms of $(1)$, we simply modify partial derivatives $\partial_{\mu}$ to $$ \partial_{\mu}U \to \partial_{\mu} - iR_{\mu}U + iUL_{\mu}, $$ where $$ L \equiv g_{2}\frac{W_{a}\tau_{a}}{2} + g_{1}W^{0}\begin{pmatrix} \frac{1}{6} & 0 \\ 0 & \frac{1}{6}\end{pmatrix}, \quad R \equiv g_{1}W^{0}\begin{pmatrix}\frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{pmatrix} $$ (in general, $g_{i}$ are matrices of constants (CKM matrix elements)).
We then may extract $\pi^{\pm}W^{\mp}$ vertex from the kinetic term of $(1)$ (here $W^{\pm} \equiv \frac{1}{\sqrt{2}}(W_{1} \mp iW_{2})$). Since, as I've written above, pion fields as the goldstone phase are always in combination $\frac{\pi}{f_{\pi}}$ and by taking into account that there are $f_{\pi}^{2}$ in the front of it, we have that $\pi W$ term is proportional to $f_{\pi}$.
Next, gauging of the Wess-Zumino term is harder (see an answer here). But now is only one important thing - the degree of $f_{\pi}$, so we immediately may give the result: the part of gauged WZ term which contains one pion field is inversely proportional to $f_{\pi}$.
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