What really is an indeterminate form? We can apply l’Hôpital’s Rule to the indeterminate quotients $ \dfrac{0}{0} $ and $ \dfrac{\infty}{\infty} $, but why can’t we directly apply it to the indeterminate difference $ \infty - \infty $ or to the indeterminate product $ 0 \cdot \infty $?
Furthermore, why can’t we call $ \infty + \infty $ and $ \infty \cdot \infty $ indeterminate forms?
I’m new to calculus, so please clear up my concepts if you can. Thanks!

Your first question is very simple: L'Hôpital's rule is a for things of the form $0/0$ and $\infty/\infty$. Therefore, if you have a problem not of these forms, you can't directly apply L'Hôpital's rule.
A better question would be to ask why L'Hôpital's rule does work for those two forms. IMO, the simplest explanation involves Taylor series and maybe Laurent series, although we can use differential approximation to get the flavor of what's going on.
We can write any differentiable function in the form
$$ f(x) = f(0) + x f'(x) + x r(x) $$
where $r(x)$ is a function (you can solve this equation for it) that satisfies
$$ \lim_{x \to 0} r(x) = 0 $$
Exercise: prove that the above is true. That is, solve for $r(x)$, and compute the limit. 
In the case that $f(0) = 0$, this takes the special case
$$ f(x) = x f'(x) + x r(x) $$
Similarly, if we have another function $g(0) = 0$, we also have
$$ g(x) = x g'(x) + x s(x) $$
and so
$$ \frac{f(x)}{g(x)} = \frac{f'(x) + r(x)}{g'(x) + s(x)} $$
If $g'(0) \neq 0$ and $f'$ and $g'$ are continuous at $0$, then we can take the limit of the right hand side without trouble.
So L'Hôpital's rule can be thought of, in this case, as a way to cancel out the zero -- i.e. to cancel out one factor of $x$ -- from both the numerator and the denominator.

To extend this more generally, suppose we can write
$$ f(x) = x^n f_0(x)  \qquad \qquad g(x) = x^n g_0(x) $$
where $g_0(0) \ neq 0$. Clearly, we have
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f_0(x)}{g_0(x)}$$
If we apply L'Hôpital's rule, we would get
$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{n x^{n-1} f_0(x) + x^n f_0(x)}{n x^{n-1} g_0(x) + x^n g_0(x)} 
$$
So long as $f'_0(x)$ and $g'_0(x)$ do not diverge to $\pm \infty$ at $x=0$, it's easy to see we get the correct limit, by first canceling out an $x^{n-1}$ from everything.
This argument doesn't work in full generality, but it wasn't meant as a proof -- it was meant to convey that L'Hôpital's rule works, by cancelling things out in some subtle sense, which is why the two forms it applies two are both quotients, where you want to cancel something out (zeroes or a poles) to find the limit.
