I am having some trouble understanding why some functions are members of $L^2(\mathbb{R})$ whereas other functions are members of some restricted subset of $\mathbb{R}$ such as $(-\pi, \pi)$
Can someone explain to me why is it true that $f(x) = \sin(x)$ an element of $L^2(-\pi, \pi)$ but not $L^2(a,b)$ for arbitrary $a,b$ defined on the real line?
This is simply false. $\sin$ is locally square integrable, i.e. $\sin \in L^2([a,b])$ for all $a<b\in\mathbb R$. This can also be written as $\sin \in L^2_{\text{loc}}(\mathbb R)$. To see that estimate $$\int_a^b \sin^2 x \ \mathrm dx \le \int_a^b 1 \ \mathrm dx = (b-a) < \infty$$ Thus $\sin \in L^2([a,b])$ for any $a<b\in\mathbb R$, so we are done.
To see why $\sin\notin L^2(\mathbb R)$ note that $\sin\ne 0$ and $\sin$ is continuous. This implies that $\|\sin\|_{L^2([-\pi,\pi])} = \epsilon > 0$. Now using periodicity we see that $$\int_{\mathbb R} \sin^2 x \ \mathrm dx = \sum_{k\in\mathbb Z} \int_{2\pi k - \pi}^{2\pi k + \pi} \sin^2 x \ \mathrm dx = \sum_{k\in\mathbb Z} \epsilon \to \infty$$ So $\sin$ doesn't belong to $L^2(\mathbb R)$.