In other words, if I have a single-variable function can I write it as a multi-variable function (without changing the mapping at all)?
I feel like the answer should be yes, because I should just be able to define $g(x,y) \equiv f(x)\quad \forall y \in Y$
If this is indeed true, then does that mean that the set $\{g\vert g:X\times Y\to \mathbb{R}\}$ nests the set $\{f\vert f:X\to\mathbb{R}\}$?
At their heart, these functions are sets themselves. A function $f$ from $A$ to $B$ is a subset of $A \times B$ with the property that, for each $a \in A$, there exists exactly one $b \in B$ such that $(a, b) \in f$. We notate this $f(a) = b$. The uniqueness of $b$ for a given $a$ means that $f(a)$ is well-defined for each $a \in A$.
In terms of what you're asking, let's take an example. Define two functions: \begin{align*} f &: \mathbb{R} \to \mathbb{R} : x \mapsto x, \\ g &: \mathbb{R} \times \mathbb{R} \to \mathbb{R} : (x, y) \mapsto x. \end{align*} Then, \begin{align*} f &= \lbrace (x, x) : x \in \mathbb{R} \rbrace, \\ g &= \lbrace ((x, y), x) : x, y \in \mathbb{R} \rbrace. \end{align*}
Note that these are not the same sets. We have, for example, $(1, 1) \in f$, but $(1, 1) \notin g$. Similarly, we have $((1, 7), 1) \in g$, but $((1, 7), 1) \notin f$. They are not literally the same function.
That said, these functions are "the same" in an intuitive sense, as you've noticed. But in terms of set theory, no, they're not the same.