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.
There are several interesting questions in the main question, plus a point in the comment that I want to address. Similar ideas are discussed here and in arxiv 0702.365.
Disclaimer : I will only speak about QFT that have a finite UV cut-off $\Lambda$. That remove all the complications of the definition of the (non-pertubative) continuum limit $\Lambda\to \infty$, as discussed in the answer linked above. It is only if you believe that a particular QFT is the absolute, true description of the universe that this limit is interesting. And it's pretty sure that it is not the case. Nevertheless, we can be interested in the limit where $\Lambda$ is very large compare to all other energy scales (which is equivalent to take $\Lambda\to\infty$).
First of all, bare parameters can be physical, and measured. They just don't correspond to the same quantities than renormalized parameters. For instance, take the (classical) Ising model. It has one coupling constant $K=J/T$. Using standard calculations, one can rewrite the partition function as a field theory with action $$ S[\phi_i]=\sum_{ij} t_{ij} \phi_i \phi_j- \sum_i \ln \cosh \phi_i, $$ where $\phi_i$ is the value of the field on a lattice site $i$, and $t_{ij}$ is related to the interaction energy $K$ (see for example this article for the details). Thus, if you know $K$, which is accessible (for instance in the case of simulations), you know the bare parameters ! You can also measure this kind of parameters experimentally (exchange energy).
Notice that this field theory is non-perturbative (if one expands the potential, all the coupling constants (and there are an infinity of them) are of the same order ! The only reason one can use the pertubative $\phi^4$ theory to describe the Ising model is because one is usually interested only in the universal quantities, that don't care about the details of the microscopic theory, as long as the universality class is the same.
For the sake of completeness : if one expand the potential (the $\ln\cosh$) to the order four, the quadratic term will be called the mass term and the fourth term the interaction. There is also a contribution to the mass coming from $t_{ii}$. One can then show that for $K$ large enough, the potential has two non trivial minimum, corresponding to the ferromagnetic phase. The critical value of $K$ for the transition, noted $K^0_c$, is at the mean-field level both wrong and non-physical.
So far so good. Now let's add the fluctuations of the field, that will "renormalize" the theory. At one loop, one sees that the quadratic term (the "mass") has a correction which is proportional to some power of the cut-off, that is, it depends on the way the regulation is made, whether the lattice is cubic or triangular, etc. Is it a problem ? Not at all. This is just telling you that the (real, physical) critical coupling $K_c$ is non-universal, it depends strongly on the microscopic details of the system. In some sense, the calculation of these "divergent" (in reality, cut-off dependent) integrals corresponds to the calculation of the critical temperature, knowing the microscopic physics.
From the Wilsonian RG, we know that some quantities will depend on the cut-off, such as the critical temperature. They are usually very hard to compute using a field theory, since the initial action is non-pertubative and one can not use the pertubative RG (wilsonian or not). Only non-pertubative schemes (numerical approaches, or the non-pertubative RG discussed in the arxiv articles linked above) can access these quantities. But there are universal quantities, such as the critical exponents, that can be computed with pertubative approach, as long as one stays in the same universality class. To compute these quantities, one has to be close to the fixed point of the RG, that is, at energy very low compare to the microscopic scales, equivalent to take $\Lambda\to \infty$.
This leads us to see the difference between Wilson RG and "old school" RG. In the former, one is imposing the microscopic value of $K$, and then look at what is the physical mass.
In the latter, one imposes the physical value of the mass, and does not care about the microscopic details, so one wants to send $\Lambda\to \infty$. One thus has to absorb the "correction" to the mass in order to fix it.
So, to (finally, but partially) answer your question "The running of the coupling in Wilson's approach has nothing to do with the bare parameters going to infinity when the cutoff is removed right?" :
In the Wilsonian approach, one starts from the microscopic scale $\Lambda$ and looks at what's going at smaller energy, whereas in the "standard" approach, one fixes than macroscopic scale and sends $\Lambda\to \infty$ in order to effectively probe smaller and smaller energy scales.
Best Answer
The subtility in the OP's question is in the use of dimensionfull and dimensionless quantities. This is easier to understand if one does not work at the upper critical dimension ($d=4$ in most standard field theories), where the interaction has dimension (even using dimensionfull quantities). I will first comment on the case where the UV cut-off $\Lambda$ is finite, and then comment on the limit $\Lambda\to \infty$.
To give an example, let us look at a massless $\phi^4$ theory in $d=5$, which would correspond to a trivial theory in a renormalization group sense, the infrared fixed point being the gaussian fixed point. This, however, does not mean that the renormalized parameter of the theory are all zero ! The most interesting quantities are low energy quantities, which correspond to the "renormalized parameters" of the OP, like the (renormalized) mass $m_R$ (here equal to zero) and the (renormalized) interaction $g_R$. Physically, $g_R$ corresponds to a scattering amplitude at zero 5-momentum $g_R$, and it is a priori finite (that is, unless the bare theory is non-interacting), as one could compute in a perturbative expansion in $g_B$, the bare value of the interaction. (Note that $g_R$ will depend on $\Lambda$, but that is not a problem, since $\Lambda$ is assumed finite.)
Now, what do we expect in a renormalization group point of view? First, we need to use dimensionless quantities, and introducing $\tilde g= \mu g$, we find that after a short transient regime (corresponding in the dimensionfull units to the transition from $g(\mu=\Lambda)=g_B$ to $g(\mu\ll\Lambda)=g_R$), the renormalized dimensionless interaction flows to zero (i.e. to the gaussian fixed point) $\tilde g(\mu)\propto \mu$, while the dimensionfull interaction is constant.
The OP was interested in the case of a relevant variable, which can be included easily in the example by assuming that the theory is massive, but with a very small (dimensionfull, renormalized) mass $m_R\ll \Lambda$. To study the RG flow, we need a dimensionless mass $\tilde m = m/\mu$, and we see that in the IR, $\tilde m$ is a relevant perturbation, that diverges as $\mu\to0$ (although, of course, the "real" mass stays finite and equal to $m_R$).
Let me now comment briefly on the issue of the limit $\Lambda\to\infty$ and the "infinite" value of the bare parameter, which is in my view an artificial problem introduced by an old-school approach to field theory (and unfortunately still promoted in textbooks and pop-science). In the present case (as well as in $d=4$), we cannot take the limit $\Lambda\to\infty$ while keeping $g_R$ finite, as there is no UV fixed point to control the flow at high-energy, and the only consistent theory is the free theory $g_B=0$. I do not really want to go further into this subtle problem (and how one could work this out $d=3$), but a related discussion is given here : Why do we expect our theories to be independent of cutoffs?