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.
Best Answer
The underlying reason for OP's flawed argument is, that a premature use of EOMs in the stationary action principle $$\begin{align} S~=~&\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \cr L(r,\dot{r};\theta,\dot{\theta})~=~&\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\end{align}\tag{A}$$ invalidates the variational principle, cf. e.g. this Phys.SE post. Concretely OP is achieving the incorrect Lagrangian (5) by eliminating the angular coordinate from the original Lagrangian (1) via the fact that the angular momentum (3) is a constant of motion (which is the EOM for the angular coordinate).
Now for OP's example, it turns out that a correct reduction can be perform via the Hamiltonian formulation $$ H(r,p_r;\theta,p_{\theta}) ~=~\frac{p_{r}^2}{2m}+ \frac{p_{\theta}^2}{2mr^2} + V(r). \tag{B}$$ Note that the canonical momentum $p_{\theta}$ (which is the angular momentum) is a constant of motion because $\theta$ is a cyclic variable.
We next re-interpret the system (B) in a rotating frame following the particle with fictitious forces and only 1D radial kinematics. The Hamiltonian (B) becomes $$ H(r,p_r)~=~\frac{p_r^2}{2m}+ V_{\rm cf}(r)+ V(r), \tag{C}$$ where $$ V_{\rm cf}(r)~:=~\frac{p_{\theta}^2}{2mr^2} \tag{D}$$ is a centrifugal potential in a 1D radial world, cf. my Phys.SE answer here. Hence the Hamiltonian system (C) has effectively only one radial degree of freedom.
Finally, perform a Legendre transformation on the Hamiltonian (C) to obtain the corresponding 1D Lagrangian system: $$ L(r,\dot{r})~=~\frac{m}{2 }\dot{r}^2\color{Red}{-}V_{\rm cf}(r)-V(r).\tag{E} $$ Note the crucial minus sign marked in red. One may check that the corresponding EL eq. for (E) is the correct EOM.
The Lagrangian (E) is minus Routhian, cf. this Phys.SE post.