Suppose that $X$ is a locally convex space with $\mathcal{E}(X)$ being the cylindrical $\sigma$-algebra.
Let $\gamma$ be a Gaussian measure on $X$. Bogachev defines the reproducing kernel Hilbert space $X_\gamma^\ast$ as the closure of the set $$ \{f - a_\gamma(f) : f \in X^\ast\}, $$ in the $L^2(\gamma)$ norm. Above, $a_\gamma(f) := \int f \, d \gamma$.
At the bottom of page 44, he seems to indicate that it is possible for $g \in X_\gamma^\ast$ to have $a_\gamma(g) \neq 0$.
However, I am confused. Suppose $\epsilon > 0$, then there is $h \in X^\ast$ with such that $\|g - (h - a_\gamma(h))\|_{L^2(\gamma)} \leq \epsilon$. This would seem to imply that $$ |a_\gamma(g)| \leq \Big|\int g - (h - a_\mu(h)) d\gamma \Big| + \Big|\int h - a_\mu(h) d\gamma \Big| \leq \|g - (h-a_\mu(h))\|_{L^2(\gamma)} \leq \epsilon? $$ This would seem to indicate that $a_\gamma(g) = 0$.
What in my reasoning is wrong here?
Your reasoning is correct, I think the confusion is caused by you misunderstanding what Bogachev is saying so let me try restating it. The quick summary is that the claim is made for $f \in X^*$ rather than $f \in X_\gamma^*$.
At this stage in that book, you have maps $R_\gamma : X^* \to (X^*)'$ defined by $$R_\gamma(f)(g) = \int (f - a_\gamma(f)) (g - a_\gamma(g)) d\gamma$$ and $\bar{R}_\gamma: X_\gamma^* \to (X^*)'$ (where I add the bar in comparison to Bogachev's notation for clarity) defined by $$\bar{R}_\gamma(f)(g) = \int f (g - a_\gamma(g)) d\gamma.$$
At the bottom of page 44, Bogachev is observing that if $f \in X^*$ then $f - a_\gamma(f) \in X_\gamma^*$ and $$R_\gamma(f) = \bar{R}_\gamma(f - a_\gamma(f)).$$ However, if $a_\gamma(f) \neq 0$ then $f \not \in X_\gamma^*$ (by the argument you include in your question) so that $\bar{R}_\gamma(f)$ is not well-defined.