Let $F:\omega^\omega\rightarrow\omega^\omega$ be a total function.
According to definitions given by Sacks (Higher Recursion Theory) and Rogers (Theory of Recursive Functions) regarding recursive subsets of $\omega^\omega\times\omega^\omega$, I gather that $F$ is said to be recursive iff $A$ $=$ $\text{graph}(F)$ $\subseteq\omega^\omega\times\omega^\omega$ is recursive. And using their definitions, I have that $A$ is recursive iff there is some $e\in\omega$ s.t. $\forall f\forall g[\chi_A(f,g)=\varphi_e^{f,g}(0)]$, where $\chi_A$ is the characteristic function of $A$.
Now, according to class notes from a Mathematical Logic course I once took and Van Wesep's informal discussion (Foundations of Mathematics pgs. 388-389 http://www.mathetal.net/data/book1.pdf), $F$ is said to be recursive iff there is some $e\in\omega$ s.t. $\forall f[F(f)=\varphi_e^f]$, where $\varphi_e^f$ is total for every $f$.
If my first interpretation of what it means for $F$ to be recursive is correct, then how are these two notions equivalent/connected? If my first interpretation is flawed, then what should I take to be the definition of what it means for $F$ to be recursive in the context of Sacks' and Rogers' books?
You must misunderstand the first one.
There is no function $F$ so that $G^F$, the graph of $F$, is recursive. Otherwise, suppose that $F(f)=g$, then $\varphi_e^{\sigma,\tau}(0)=1$ for some $e$ and finite strings $\sigma\prec f$ and $\tau\prec g$. Then for any $g_0\succ \tau$, $(f,g_0)\in G^F$. So $F$ cannot be a function.