The easiest way, I think, is by using the spurion analysis. If an operator $\mathcal{O}$ with a small coupling $g$ in the lagrangian breaks a symmetry, you just need to promote $g$ to a field (called spurion) with definite transformation properties such that $g\mathcal{O}$ is now invariant. The low-energy effective action now must be written making an expansion in $g$ by writing down all the invariants to any given order, and eventually freeze $g$ to its original value.
Let me give you the example of chiral $SU(2)$ in QCD. In this theory there is an approximate $G=SU(2)L\times SU(2)_R$ symmetry of the UV theory for up and down quarks $$ Q{L}=\left(\begin{array}{c}u_L\ d_L \end{array}\right)\rightarrow U_L Q_{L}\,,\quad Q_{R}=\left(\begin{array}{c}u_R\ d_R \end{array}\right)\rightarrow U_R Q_{R} $$ where $U_{L,R}$ are unitary $2\times 2$ matrices of $SU(2){L,R}$. In the IR, the symmetry is broken spontaneously down to a subgroup $H=SU(2){L+R}$ isospin so that 3 massless pions $\pi^a$ are present in the low-energy spectrum according to the <PERSON> theorem. This pions can be written in a non-linear sigma model fashion as $\Sigma(x)=e^{i\pi^a(x) \sigma^a/f_\pi}$ where $\Sigma\rightarrow U_L \Sigma U^\dagger_R$ non-linearly realize $G$ but linearly realize $H$ (where $U_L=U_R$). The lowest order lagrangian invariant under $G$ is of the schematic form $$ \mathcal{L}^{(2)}{eff}=\frac{f\pi^2}{4}\mathrm{Tr}\left[\partial_\mu\Sigma^\dagger \partial^\mu\Sigma\right]. 976
$$ Now, we add a small explicit breaking of $G$ in the UV, namely a mass term $m_u \bar{u}L u{r}+m_d \bar{d}L d{r}+h.c.$ which we can write as $$ \bar{Q}{L}M Q{R}+h.c.\,,\qquad M=\left(\begin{array}{cc} m_u & 0 \ 0 & m_d\end{array}\right)\,. $$ In fact, this mass term doesn't even respect $H$ (unless $m_u\neq m_d$) and not only the pions will get mass but they will also be split. Anyway, the UV lagrangian can be thought as invariant under $G$ if we promote $M$ to a spurion and transform it as $M\rightarrow U_L M U^\dagger_{R}$. Now, going in the IR and looking at the effective theory of the pions, we can write new invariants in the action for $\pi$. The most important are the ones with the least insertions of $M$ since we assumed that the explicit breaking was small (in this case, compared to $\Lambda_{QCD}$). There is only one operator with just one insertion of $M$, namely $$ \delta\mathcal{L}=c\mathrm{Tr}\left[\Sigma^\dagger M+\Sigma M\right]\,. $$ Now one can freeze $M$ to $M=\mathrm{diag}\left(m_u,m_d\right)$, expand $\delta\mathcal{L}$ in powers of the pion fields, and thus get the form of the potential (in particular the mass terms) at leading order in the explicit breaking parameter $M$. Just for simplicity, assume now that the explicit breaking of $G$ respects $H$, $m_u=m_d\equiv m$, so that $M=m\mathbf{1}$ is multiple of the identity. 818