I gave a much more detailed answer at
What is the Wilsonian definition of renormalizability?
so try to read it.
But a quick explanation is as follows: Let $\Lambda=2^n$ be your UV cutoff.
You are dealing with a sequence $n=1,2,\ldots$ of bare pure $\phi^4$ theories which are starting points for RG transformations, say a transformation $T$ corresponding to zooming by a factor of 2. The effective theory at say unit scale is the limit of $T^n$ applied to the starting point corresponding to $\Lambda=2^n$, when $n$ goes to infinity. So the starting points are pure (no $\phi^6$ junk etc.) but the effective theories will usually contain in the Lagrangian all terms allowed by the symmetries.
Edit: The OP asks me to point out where the OP's confusion is, so I'll try.
I will use the notations of my long answer linked to above. The UV cut-off in momentum space is $\Lambda=e^{-t}$ where $t$ will eventually be taken to $-\infty$ in order to perform the construction of a continuum QFT. The bare action is
$$
V_{t}^{\rm bare}(\phi)=\sum_{i\in I}g_i^{\rm bare}(t)\int \mathcal{O}_{i}(x)\ d^dx
$$
for Wick-ordered local operators
$$
\mathcal{O}_{i}(x)=:\partial^{\alpha_1}\phi(x)\cdots\partial^{\alpha_k}\phi(x):_t
$$
and bare couplings $g_i^{\rm bare}(t)$. Here $I$ is just an index set of labels for operators. Note that Wick-ordering is with respect to the Gaussian theory with cut-off at scale indicated by $t$, hence the $:\cdots:_t$
notation.
Regardless of one's choice of method, modern Wilsonian RG or old BPHZ counterterm renormalization, what needs to be done to construct the continuum QFT is to choose the right ansatz for the bare coupling dependence on the UV cut-off, i.e., the functions $g_i^{\rm bare}(t)$, so that
$$
\forall t_2, \ \lim_{t_1\rightarrow -\infty}\ RG[t_2,t_1](V_{t_1}^{\rm bare})=:V_{t_2}^{\rm eff}\ {\rm exists}.
$$
Here $RG[t_2,t_1](V_{t_1}^{\rm bare})$ is the (temporary) effective theory at (log, position space) scale $t_2$ for the bare theory with cut-off at scale $t_1$. What matters in the end is $V_{t_2}^{\rm eff}$ which is the effective theory at scale $t_2$ of the wanted continuum QFT.
The OP's confusion is to think
that $V_{t}^{\rm bare}$ and $V_{t}^{\rm eff}$ are the same when they are not.
If one is constructing a $\phi^4$ theory, this means that
the only nonzero bare couplings $g_i^{\rm bare}(t)$ one is allowed to play with are the couplings for the operators $(\partial\phi)^2$, $\phi^2$ and $\phi^4$. It is fine to transgress this rule, but then one would be constructing some other theory like a $\phi^6$ model, etc.
Now even if one adheres to the rule and succeeds in removing the cut-off, the effective theories
$$
V_{t}^{\rm eff}(\phi)=\sum_{i\in I}g_i^{\rm eff}(t)\int \mathcal{O}_{i}(x)\ d^dx
$$
will have nonzero couplings for pretty much all operators allowed by symmetry. This is because the RG is a highly nonlinear transformation.
Finally, the long answer I gave is more adapted to the generic situation (like $\phi^4$ in 3D) where the perturbed CFT has finitely many relevant operators. The 2D situation mentioned by the OP is special: all powers $\phi^k$ are relevant at the Gaussian fixed point. So the procedure to find the right bare ansatz $g_{i}^{\rm bare}(t)$ is different. The easiest choice which works is to take $g_{i}^{\rm bare}(t)$ constant with respect to the UV scale $t$.
All the counterterm renormalizations (just tadpoles) are hidden in the Wick-ordering $:\cdots:_t$.
