When we have a function $f: \mathbb{R} \to \mathbb{R}$, I can intuitively picture that and think that for every $x \in \mathbb{R}$, we can find a $y \in \mathbb{R}$ such that our function $f$ maps $x$ onto $y$.
I'm confused, however, when we have something like: $g: D \to \mathbb{R}$, where $D$ is the domain of our function such that $D \subset \mathbb{R}$. How can our function output every element in $\mathbb{R}$, when our input was specifically less than the whole set of reals?
Perhaps I'm understanding your question differently than some of the other commenters, but I'll point out that saying you have a function $g: D \to \mathbb{R}$ does NOT mean every element of $\mathbb{R}$ gets outputted. Here, the set that comes after the arrow is something called a codomain which can be larger than the range. If every element of your codomain gets outputted, then your function has a specific property called surjectivity, but not every function will have this. For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. Here, $\mathbb{R}$ is NOT the range of our function since there is no real number that maps to the number $-1 \in \mathbb{R}$ under $f$.
I don't agree with your statement that the range is a larger set than the domain.
I give some counter-examples to illustrate.
If you take any constant function defined as $f: R \rightarrow R$, $f(x) \equiv k$ (fixed value of $k$), then
Domain$(f) = R$
Range$(f) = \{ k \}$
As another example, take the 'sign' function defined as
$g: R \rightarrow R$, where $g(x) = \mbox{sign}(x)$.
Then
Domain($g$) = $R$
Range($g$) = $\{ -1, 0, 1 \}$
(It is a convention to define sign(0) = 0.)
If you define $h: R \rightarrow R$ as $h(x) = | x |$, the absolute value of $x$, then
Domain($h$) = $R$
Range($h$) = non-negative reals, which is a smaller set than $R$.
The underlying idea of your question is the concept of cardinality. We say that two sets $X$ and $Y$ have the same cardinality iff the exists a bijection $f:X\to Y$ between them. Such relation is an equivalence relation, that is to say, it is reflexive, symmetric and transitive. Hence each cardinality is an equivalent class. We can establish a linear order relation among the possible cardinalities. Finally, we say that that a set $X$ is infinite iff there exists a bijection between $X$ and a proper subset of it.
For example, the set of natural numbers is infinity because $f:\textbf{N}\to P$, where $f(n) = 2n$ is a bijection between $\textbf{N}$ and the set $P := \{n\in\textbf{N} \mid (n = 2k)\wedge(k\in\textbf{N})\}\subset\textbf{N}$.
At the given example, one can consider the function $\tan:(-\pi/2,\pi/2)\to\textbf{R}$ which is bijective (as suggested by @MichaelMorrow). Therefore, even though the interval $(-\pi/2,\pi/2)$ is a proper subset of $\textbf{R}$, they have the same cardinality, hence the set of real numbers $\textbf{R}$ is infinity.
One interesting result is that $\operatorname{card}(\textbf{N}) < \operatorname{card}(\textbf{R})$, but it is not known if there is a cardinality between both. Cantor has conjectured it and such statement is known as the continuum hypothesis.
If $A$ is a finite set, then it is not possible for there to be a function $f:A\to B$ such that the range is a proper superset of the domain. However, this statement is not true for infinite sets, which is just one reason why they are so counter-intuitive. For instance, we can construct a surjective function $f:\mathbb Z^+\to\mathbb Z$ by mapping the odd positive integers to the nonnegative integers, and mapping the even positive integers to the negative integers. Similarly, the function $f:(-\pi/2,\pi/2)\to\mathbb R$ given by $f(x)=\tan x$ is surjective, even though $(-\pi/2,\pi/2)$ is clearly a proper subset of $\mathbb R$.
