Consider the following Lagrangian describing the interaction between a massless field $\phi$ and a massive field $\psi$: $$ {\scr L} = \frac12(\partial\phi)^2 + \frac12 (\partial\psi)^2(1 + f(\phi/M)) - \frac12m^2\psi^2.\tag{1} $$ The interaction between these two fields will generate an effective potential for $\phi$, which we can compute using the background field method. First we shift $$\phi\to\phi_b + \delta\phi\tag{2},$$ with $\phi_b$ constant $$ {\scr L} = \frac12(\partial\delta\phi)^2 + \frac12(\partial\psi)^2(1 + f[(\phi_b+\delta\phi)/M]) - \frac12 m^2\psi^2.\tag{3} $$ We can canonically normalize $\psi$ by rescaling: $$\psi_C = \sqrt{1 + f(\phi_b/M)}\psi\tag{4},$$ yielding $$ \frac12(\partial\delta\phi)^2 + \frac12(\partial\psi_C)^2\left(1 + \sum_{n = 1}^\infty \frac{f^{(n)}(\phi_b/M)}{(1 + f(\phi_b/M))n!} \left(\frac{\delta\phi}{M}\right)^n\right) - \frac12 m^2[\phi_b]\psi_C^2,\tag{5} $$ where $$ m^2[\phi_b] = \frac{m^2}{1 + f(\phi_b/M)}.\tag{6} $$ We can compute the path integral over $\psi$ in the gaussian approximation by dropping the 3 and higher-point interactions multiplying $(\partial\psi_C)^2$. The result is standard (see, e.g. Schwartz page 745 and surrounding discussion) $$ V_{\rm eff}^{1\,\rm loop} = \frac{-i}{2}\int\frac{{\rm d}^4 k}{(2\pi)^4}\log\left(1 - \frac{m^2[\phi_b]}{k^2}\right).\tag{7} $$ Wick rotating, we find $$ V_{\rm eff}^{1\,\rm loop} = \frac{1}{64 \pi ^2} \left(\Lambda ^2 m\left(\phi _b\right){}^2-m\left(\phi _b\right){}^4 \log \left(\frac{\Lambda ^2}{m\left(\phi _b\right){}^2}+1\right)+\Lambda ^4 \log \left(\frac{m\left(\phi _b\right){}^2}{\Lambda ^2}+1\right)\right),\tag{8} $$ where $\Lambda$ is a Euclidean cutoff on the loop momentum. At leading order, this result is quadratically divergent.
We can cross-check this answer by comparing $V_{\rm eff}''(\phi_b)$ to the self-energy diagram evaluated at zero external momentum. At one loop, there are two such diagrams, coming from the $(\partial\psi)^2\delta\phi$ and $(\partial\psi)^2\delta\phi^2$ interactions. Now comes the confusion: one can read off from these interactions that the resulting self-energy diagram contains a quartic divergence as opposed to a quadratic divergence. One may hope that these divergences cancel precisely, but this is not the case for general $f(\phi/M)$: $$ m_{\phi,{\rm eff}}^2 = \frac{\Lambda^4}{64\pi^2 M^2}\left(\frac{f''(\phi_b/M)}{1 + f(\phi_b/M)} - \frac{{f'(\phi_b/M)}^2}{(1 + f(\phi_b/M))^2}\right)+ V_{\rm eff}''(\phi_b).\tag{9} $$ I find this result extremely strange, since I was under the impression that it was a theorem that the effective potential generated all 1PI diagrams. Further, the fact that the results agree precisely up to the quartic divergence is further evidence to me that something pathological is going on.
One could resolve this tension by taking $f(\phi/M) = \exp(\lambda\phi/M) - 1$, but I hardly find this convincing. Another possible resolution is to canonically normalize $\psi$ before shifting to the background field, thus eliminating all vertices involving $\psi$-momentum. However, now the self-energy of $\psi$ contains a quartic divergence.
I will finally note that this problem is not present in dimensional regularization, since dim reg automatically drops polynomial divergences. However, if one is concerned about the UV sensitivity of the model, the cutoff scale $\Lambda$ is physical, and the quadratic vs quartic divergence dramatically changes the character of the effective potential.
