About question 2: I think that the software Singular has this feature; it's well-documented, and if you look for resolution graph you should find the reference.

About question 1: well, I must admit that that algorithm is not pleasant, and that it took me a while to work out the case of $\Sigma(2,3,4)$.

Anyway, here we go. I can't draw graphs here (but if anyone can tell me that we can embed \xygraphs, next time I can make an effort), so things will have to be less pictorial than I'm comfortable with. I'll use greek letters for vertices to avoid clashing with Némethi's letters. So, we fix $g(x,y) = x^2+y^3$ and $n=4$.

First off, the resolution graph of $g$ has three vertices $\alpha, \beta, \gamma$, and an arrowhead $\delta$. The self-intersections are $e_\alpha = -3$, $e_\beta=-2$, $e_\gamma = -1$. ($\gamma$ is the central vertex, corresponding to the last blow-up, $\alpha$ was the first blow-up, and $\beta$ the second.)

From the formula (*) in the appendix, relating multiplicities with self-intersections, and using that $m_\delta = 1$, we obtain that $(m_\alpha, m_\beta, m_\gamma) = (2,3,6)$. We the compute $(d_\alpha, d_\beta, d_\gamma) = (2,1,3)$.

Step 1(a): from the computations above, $\alpha$ is covered by two vertices of multiplicity 1, while $\beta$ and $\gamma$ are covered by one vertex of multiplicity 3 each. Call them $\tilde\alpha_1, \tilde\alpha_2, \tilde\beta, \tilde\gamma$.
The genus formula says that they're all rational ($\tilde g = 0$).

Step 1(b): the edge $(\alpha,\gamma)$ lifts to strings connecting $\tilde\alpha_i$ to $\tilde\gamma$, each of type $G(1,3,2)$; which means just a single vertex $\varepsilon_i$ with $(e_{\varepsilon_i}, m_{\varepsilon_i}) = (-2,2)$ (and two arrowheads ending at $\tilde\alpha_i$ and $\tilde\gamma$). The edge $(\beta,\gamma)$ lifts to a string of type $G(6,3,4)$, which is a single vertex $\zeta$ with $(e_{\zeta}, m_{\zeta}) = (-2,3)$ (and two arrowheads ending at $\tilde\beta$ and $\tilde\gamma$.

Step 1(c): the arrowhead $\delta$ lifts to a string of type $G(6,1,4)$, which is again a single vertex $\eta$ with $(e_{\eta}, m_{\eta}) = (-2,2)$ (and two arrowheads, one ending at $\tilde\gamma$ and the other one free).

At this point, the graph is star-shaped with four legs: its center is at $\tilde\gamma$ (with weight unspecified); two legs with weights $(-2,?)$ (ending at $\tilde\alpha_1, \tilde\alpha_2$), one with weights $(-2,?)$ (ending at $\tilde\beta$), one with weight $-2$ and an arrowhead.

Step 2: we now compute the missing self-intersections using (*). We obtain $e_{\tilde\alpha_i} = -2$, $e_{\tilde\beta} = e_{\tilde\gamma} = -1$.

Step 3: we drop the arrowhead, and blow down $\tilde\gamma$ and then $\zeta$ (which, after blowing down $\tilde\gamma$, can be blown down). The graph we obtain is $E_6$, as expected.