I assume/hope the following post should clarify the issue: I'm missing the point of renormalization in QFT.
We don't use two different coupling constants, we use the same constant.
The perturbative expansion is always in terms of
$$
\lambda:=G_4(p)|_{p^2=\mu^2}
$$
where $G_4=\langle\phi\phi\phi\phi\rangle$ is the four-point function and $\mu$ is some energy scale of interest. You can choose whatever $\mu$ you want, typical choices are $\mu=0$, $\mu=m$ (the pole mass), or $\mu=\sqrt s$, where $s$ is the energy of whatever process you are measuring.
Perturbative predictions always take the form
$$
A=a_0+a_1\lambda+a_2\lambda^2+\cdots
$$
where $a_i$ are numerical coefficients. This is the expression that you share with your experimental friend in order to compare to their measurements.
Note that $\lambda$ is not a parameter that appears in your Lagrangian. If you let $\lambda_0$ be the coefficient of $\frac{1}{4!}\phi^4$ in your Lagrangian, then you can write $\lambda=\lambda_0+c_1\lambda_0^2+c_2\lambda_0^3+\cdots$ for some coefficients $c_i$. Of course, given that $\lambda=\lambda_0+\cdots$, you can also express the perturbative expansion of $A$ as a series in $\lambda_0$. But this is not useful, since $\lambda_0$ is unmeasurable by itself. On the other hand, $\lambda$ can be measured in the lab, and hence it is much more convenient to express predictions as a series in $\lambda$.
If you only care about tree-level calculations, you can take $\lambda=\lambda_0$, and forget about the fact that $\lambda$ and $\lambda_0$ are two different objects. But if you want to look at higher order contributions, then you have two different objects, and the distinction becomes important.
