We start with a Lagrangian full of bare couplings and fields $\{Z_B,g_B,m_B,...\}$. Then we regularize the theory by a hard cut-off $\Lambda $ or by dimensional reg $d=4-\varepsilon$ to tame the infinities. The amplitudes can then be formally calculated in perturbative expansion on the bare coupling $g_B$:
$$
\mathcal{M}(s_0,t_0,u_0)=\sum_{n=1}^{\infty}A_n(Z_B,m_B,...,\Lambda\, or\,\varepsilon)(g_{B})^{n}
$$
Each term $A_n(Z_B,m_B,...\Lambda\, or\,\varepsilon)$ usually diverge as we make $\Lambda\rightarrow\infty$ or $\varepsilon\rightarrow 0$, so this can be viewed only as formal expression. The idea of renormalizability is to absorb these divergence doing bare couplings $\rightarrow$ renormalized couplings.
Now, that the infinities are tamed as long as we hold $\varepsilon\neq0$ or $\Lambda\neq\infty$, we can do the bare couplings $\rightarrow$ renormalized couplings (i.e. fixing a renormalization condition).
$$
\mathcal{M}(s_0,t_0,u_0)=g_0
$$
for a given renormalization scale $\mu_0$ contained in $(s_0,t_0,u_0)$. Together with others renormalization conditions we obtain a set of renormalized couplings $\{Z_0,g_0,m_0,...\}$. Note that this depends on the regularization procedure and the scale $\mu_0$. This renormalized couplings are finite and can be obtained directly or indirectly by experimental results.
The running coupling is the $\{Z(\mu),g(\mu),m(\mu)\}$, and can be obtained by perturbative calculation, expanding the beta functions in the renormalized couplings $\{Z_0,g_0,m_0,...\}$. Note that the running of the couplings are telling you how the renormalization conditions should change if you change the scale by $\mu_0\rightarrow\mu$ in order to preserve all the physical quantities (preserve the predictions of the theory).
Now, using the renormalization condition $\mathcal{M}(s_0,t_0,u_0)=g_0$ and the formal expansion of $\mathcal{M}(s_0,t_0,u_0)$ in terms of the bare couplings you can relate the bare and the renormalized couplings. Knowing the beta-functions you can relate the renormalized couplings of two diferent scales $\mu_0$ and $\mu$ of the same theory.
So, all this relate the bare couplings, the renormalized couplings and the running couplings.
Now, the Wilson approach is somehow different. We start with a theory that is already regulated by a cut-off $\Lambda_0$ and the couplings can be viewed as functions of the cut-off $\{Z(\Lambda_0),g(\Lambda_0),m(\Lambda_0),...\}$. As we integrate over modes, changing $\Lambda_0\rightarrow\Lambda$ we obtain an effective Lagrangian with effective couplings $\{Z(\Lambda),g(\Lambda),m(\Lambda),...\}$.
Note that now there is no renormalization condition and no renormalization scale $\mu_0$. The couplings are running with the scale of the cut-off regularization instead of the renormalization scale. The relation between these two approach are only achieved when we do the limit $\Lambda \rightarrow \infty$ on both. Now, using some physical quantity in both sides we may relate the effective couplings $\{Z(\Lambda),g(\Lambda),m(\Lambda),...\}$ with the ruining couplings $\{Z(\mu),g(\mu),m(\mu)\}$.
What is reveled trough this relation is that both approaches are parametrizing the renormalization group of a "same theory", with different parametrization (is the same theory only after the limit $\Lambda\rightarrow\infty$):
- The first one is parametrizing the renormalization group by the renormalization scale $\mu$ and a renormalization condition $\{Z(\mu),g(\mu),m(\mu)\}$ and push the regularization to infinity $\Lambda\rightarrow\infty$ at the end.
- The Wilsonian approach is parametrizing the renormalization group by the regularization scale $\Lambda$ and effective couplings $\{Z(\Lambda),g(\Lambda),m(\Lambda),...\}$, and then at the end of calculation we push $\Lambda\rightarrow\infty$ as before.
Note that the effective coupling will diverge as $\Lambda\rightarrow\infty$, for the same reasons as the bare couplings does, in order to have finite physical quantities as predictions.
What is important to note is that running $\mu$ with $\Lambda$ fixed is the same as running $\Lambda$ with $\mu$ fixed. When we do $\Lambda\rightarrow\infty$, in both cases, the dependence on the the details of each regularization and approach will drop out. For more details about the relation between this two approachs see this