= Solution
The additional six-leg <interaction vertex> has <Feynman rule> $-ig_6$. Since both interactions lack derivatives, the graph identity still gives $\omega=2(1-V)$, with $V=V_4+V_6$. The difference is that a one-vertex contraction of the <phi-six theory> can leave four external legs: choosing the contracted pair gives $\binom62=15$. Its <quartic interaction> correction is
$$
\frac{g_6}{6!}\,15T_\Lambda\phi^4
=\frac1{4!}\frac{g_6T_\Lambda}{2}\phi^4.
$$
Consequently a one-loop <quartic interaction> <counterterm> is now necessary:
$$
\boxed{\delta g_4^{(1)}=-\frac{g_6}{2}T_\Lambda
=-\frac{g_6}{8\pi}\log\left(1+\frac{\Lambda^2}{m^2}\right).}
$$
The one-loop <mass counterterm> from <1/c> is unchanged. Contracting four legs of a single six-leg <interaction vertex> gives $\binom62\,3=45$ double contractions, hence a two-loop raw <self-energy> $g_6T_\Lambda^2/8$. Its subtraction must include the <quartic interaction> <counterterm> insertion, rather than subtracting this raw expression in isolation.
All divergent field-dependent coefficients are most simply organized by the <normal-ordering subtraction scheme in two dimensions>. With $T=T_\Lambda$, express a potential normal ordered at mass $m$ in ordinary fields:
$$
V_B(\phi)=\frac{g_4}{24}(\phi^4-6T\phi^2+3T^2)
+\frac{g_6}{720}(\phi^6-15T\phi^4+45T^2\phi^2-15T^3).
$$
Thus its divergent subtraction coefficients, in this scheme, are
$$
\boxed{\delta g_4=-\frac{g_6T}{2},\qquad
\delta_m=-\frac{g_4T}{2}+\frac{g_6T^2}{8},\qquad
\delta g_6=\delta_Z=0.}
$$
Indeed the two-loop <self-energy> terms are $g_6T^2/8+\delta g_4T/2=-g_6T^2/8$, which the displayed <mass counterterm> cancels. The constant terms only affect <vacuum energy>. After removing all one-vertex self-contractions, every remaining subgraph is <ultraviolet finite>. Additional finite <on-shell renormalization scheme> adjustments are subject to the same qualification as <1/d>.
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