In many textbooks, the Wilsonian and old-fashioned views of renormalization are treated as totally separate, but they are actually very closely connected. I will use your notation and describe three views of renormalization as would be taught in the three semesters of a typical QFT course.
QFT I: bare perturbation theory
Bare perturbation theory is the way renormalization is first explained in many textbooks, and the way it was first worked out. For concreteness, suppose $\phi^3$ theory is relevant to our world, and we observe that the particles have physical mass $m_p$ and physical interaction strength $g_p$, by measuring cross sections. That is, naively we have a theory with Lagrangian
$$\mathcal{L}_{\text{naive}} = (\partial \phi)^2 + m_p^2 \phi^2 + g_p \phi^3$$
where I suppress all numerical coefficients. Then the tree-level predictions of this theory roughly match that of observations. But this agreement is an illusion. When we go to one-loop order, we see the physical mass and interaction strength are instead predicted to be infinite.
In order to fix this problem, we must impose a cutoff $\Lambda$ and instead use the "bare" Lagrangian
$$\mathcal{L}_{\text{bare}}(\Lambda) = (\partial \phi)^2 + m(\Lambda)^2 \phi^2 + g(\Lambda) \phi^3$$
where $m(\Lambda)$ and $g(\Lambda)$ are formally divergent quantities, fixed so that physical predictions are finite. Concretely, for example, we may calculate the correlator $\langle \phi \phi \phi \rangle \sim g$ as a power series in $g(\Lambda)$, giving a series expansion for $g$ in terms of $g(\Lambda)$. We then flip this around to fix $g(\Lambda)$ in terms of $g$. In this way, we get finite predictions that match experiment.
QFT II: on-shell renormalized perturbation theory
Bare perturbation theory is a bit unsatisfactory. For example, it works perturbatively in $g(\Lambda)$, a formally divergent quantity. And the logic is not in the right order: we shouldn't change the theory we're working with once we see our naive theory doesn't work, we should just work with the correct theory from the start.
Instead, in renormalized perturbation theory, we start with the correct Lagrangian and split it as
$$\mathcal{L}_{\text{bare}}(\Lambda) = \mathcal{L}_{\text{naive}} + \mathcal{L}_{\text{CT}}(g_p, \Lambda).$$
Here $\mathcal{L}_{\text{naive}}$ is called the renormalized Lagrangian because it contains renormalized fields and couplings, such as $g_p$. We then perform perturbation theory in $g_p$, treating the counterterms as $O(g_p)$ and higher, which makes much more sense. The conditions used to determine the counterterms are the same as in bare perturbation theory.
To define a theory, it is sufficient to specify $\mathcal{L}_{\text{bare}}(\Lambda)$. On-shell renormalized perturbation theory, which splits this Lagrangian, is an extra layer of structure. The splitting is arbitrary, and in the case of the on-shell scheme is conceptually useful because it allows us to work with physical quantities like $m_p$ and $g_p$ throughout the calculation.
It is often said, incorrectly, that "the renormalized Lagrangian is found by adding counterterms to the bare Lagrangian". That is incorrect, because the bare Lagrangian is the whole Lagrangian; we don't add anything to it. Making this mistake causes the names to be swapped, which generates a lot of confusion.
QFT III: Wilsonian renormalization
In the Wilsonian picture, we get the best of both worlds: the naive directness of bare perturbation theory, and the proper setup of renormalized perturbation theory.
In this setup, we imagine we are performing experiments near, but below some energy $\mu$, and find particles with mass $m_p$ and interaction strength $g_p(\mu)$. We may describe these results with a Wilsonian effective action
$$\mathcal{L}_{\text{eff}}(\mu) = \mathcal{L}_{\text{naive}}|_{g = g_p(\mu)}$$
which is a reasonably good description, even when all quantum/loop effects are accounted for, because the loops get cut off at the low scale $\mu$. Thus in the Wilsonian picture observed quantities get translated directly into couplings.
Next, because we are high energy physicists, we want to use this information to find a more fundamental theory, valid up to some higher scale $\Lambda$. Let the Lagrangian for this fundamental theory be $\mathcal{L}_{\text{fund}}(\Lambda)$. Then we have
$$\mathcal{L}_{\text{fund}}(\Lambda) = \mathcal{L}_{\text{eff}}(\mu) + \Delta \mathcal{L}$$
where $\Delta \mathcal{L}$ is found by integrating out degrees of freedom between $\mu$ and $\Lambda$. Now, $\mathcal{L}_{\text{CT}}$ is found by computing the exact same integrals, but between $0$ and $\Lambda$. (The difference in the lower bound is just because the naive Lagrangian accounts for no quantum effects at all, while $\mathcal{L}_{\text{eff}}(\mu)$ accounts for quantum effects up to scale $\mu$, and don't matter all that much.)
Therefore, by comparison with what we found earlier,
- the renormalized Lagrangian is the Wilsonian effective Lagrangian
- the bare Lagrangian is the fundamental Lagrangian
- the counterterm is the term needed to compensate for the RG flow between them
Note that in all three versions presented above I've included a finite cutoff. If a continuum limit exists, we may take $\Lambda \to \infty$ for the bare/fundamental Lagrangian.
One final subtlety: when we can take this limit, the counterterms are finite! They are just the difference between the low-energy effective theory and some RG fixed point. We only think counterterms diverge because they diverge order by order in a series expansion. This doesn't mean the whole counterterm diverges; note that
$$\lim_{x \to \infty} \exp(-x) = 0$$
but the terms in the Taylor series diverge individually.
The main point that you have to always keep in mind is that relevant/irrelevant coupling constants are defined with respect to a fixed point. The standard/naive power counting is done assuming that the fixed point controlling the RG flow is the gaussian. This is true for massless QED and $\phi^4$ theories in four dimensions in the infrared, and for QCD in the UV.
However, this is not true in the opposite limits ! QED and $\phi^4$ do not have a UV fixed point, meaning the the theories are not asymptotically safe, and all 'irrelevant' coupling constants grow in the UV. On the other hand, QCD flows to strong coupling in the IR, though it seems that it still flows toward another fixed point.
All in all, what it means is that if the theory flows to a fixed point in the IR/UV, only the few relevant operators have to be fixed in order to describe all the physics at lower/higher energy.
In the case of $\phi^4$ in $d=4$, the mass is a relevant operator at the gaussian fixed point. It means that as one integrates more and more degrees of freedom, the mass drives the system away from the gaussian fixed point (it introduces a length scale in the system). But in order for the system to approach the gaussian fixed point, the mass needs to be small (otherwise, the system will go away from the gaussian fixed point before it even approaches it). This means that under the RG transformation, it will take a long time to have the mass grow up to the cut-off scale (at which point the RG flow stops).
The best reference I know that discusses the relationship between the perturbative RG (as done in QED) and the Wilsonian point of view is given in this review on the (non perturbative) RG, and in particular the section 2.6 "Perturbative renormalizability, RG flows, continuum limit, asymptotic freedom and all that".
Best Answer
At least at the operational level, whether an operator is relevant or irrelevant (in the IR) tells you about it's canonical scaling dimension.
I think the BPHZ picture of renormalization might help here. Given a physical theory, we'd like to estimate which Feynman graphs of that theory diverge. You can estimate a "superficial degree of divergence" for each Feynman diagram by considering the canonical scaling dimension of each vertex or edge in the diagram. Say you have a diagram with a vertex corresponding to an IR-irrelevant (UV-relevant) interaction term. Then, in the UV (for large momenta), it will cause the amplitude to diverge. Now, as you go to higher loop order, or insert more of those vertices at same loop order, the divergence becomes worse. And as you add more and more such vertices, the corresponding diagrams become more and more important in the UV, due to the couplings being UV-relevant! The essence is that there are infinitely many ("independent") Feynman diagrams which diverge, so you can't cook up enough counterterms to cancel all these divergences. So these terms in the lagrangian give rise to non-renormalizable theories.
See, for eg. Peskin & Schroeder section 10.1, or the link above for how to calculate the superficial degree of divergence of any diagram. Another technicality is that you don't actually consider Feynman graphs, you instead consider subgraphs. What that means is that the external (uncontracted) legs can be off-shell, as if they were sitting inside a bigger Feynman diagram.
Given your question, maybe you've already come across the stuff I've said and are asking for something else, in which case your question is not clear to me.