Understanding proof of upper bound on complexity of recursive computation of graph chromatic polynomial
You are correct that a term was omitted in the RHS of (2.3.10), but the final asymptotic bound $O(\phi^n)$ holds for the augmented RHS also. Perhaps it might be easier for you to see this if you just work out the solution for the nonhomogeneous recurrence relation from scratch rather than use Wilf's Theorem 1.4.1 on the epsilons, as shown below.  Essentially, we are  adding a linear term $an+b$ to the exponential $\phi^n$, which gives $\phi^n + an+b = O(\phi^n)$, where $\phi$ is the golden ratio.

More specifically, suppose that before the two recursive function calls to the chrompoly function, the edge set needs to be modified, and for the given value of $\gamma = |V|+|E|$, the cost for modifying the edge set is $\gamma$.  (You can also take this cost to be $c \gamma$ for some other $c$, if you wish.)

Then, the recurrence $h(n) = h(n-1)+h(n-2)+n$ needs to be solved.  From the theory of linear nonhomogeneous recurrence relations, the solution $h(n)$ is the sum of a generation solution and a particular solution, where (1) the general solution has the form $h_1(n) = c_1 \alpha^n + c_2 \beta^n$, where $\alpha , \beta$ are roots of the characteristic equation (one of them containing the golden ratio), and (2) the particular solution has the form $h_2(n) = an+b$ because the nonhomogeneous part is polynomial of degree $1$. (If the nonhomogeneous part was $7n^2 3^n$, then the particular solution would take the form $(an^2+bn+c)3^n$. If $3$ was a single root of the characteristic equation, then the particular solution would take the form $n(an^2+bn+c)3^n$. If $3$ appeared as a root with multiplicity $m$, then the extra factor would be $n^m$.)  

Thus, $h(n) = O(c \phi^n +an+b) = O(\phi^n)$ because the linear terms grow more slowly than the exponential and can be ignored in asymptotic notation.