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!
The phrase “indeterminate form” is used to mean a function that we can't compute the limit of by simply applying some general theorem.

One can easily show that, if $\lim_{x\to x_0}f(x)=a$ and $\lim_{x\to x_0}g(x)=b$, then
$$
\lim_{x\to x_0}(f(x)+g(x))=a+b
$$
when $a,b\in\mathbb{R}$. One can also extend this to the case when one or both $a$ and $b$ are infinite:


If $a=\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=\infty$ and $b=\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=-\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$
If $a=-\infty$ and $b=-\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$


(Note: $a$ and $b$ can be interchanged; $x_0$ can also be $\infty$ or $-\infty$.)

However, it's not possible to extend this to the case where $a=\infty$ and $b=-\infty$ (or conversely). We summarize this statement by saying that $\infty-\infty$ is an indeterminate form.

For instance, 


if $f(x)=x$ and $g(x)=1-x$, then clearly $\lim_{x\to\infty}(f(x)+g(x))=1$
if $f(x)=x^2$ and $g(x)=x$, then $\lim_{x\to\infty}(f(x)+g(x))=\infty$


Other cases are possible.

There are similar criterions for functions of the form $f(x)/g(x)$; if the limits of the two functions exist, then we can easily say something about the limit of the quotient, except in the cases when


$\lim_{x\to x_0}f(x)=0$ and $\lim_{x\to x_0}g(x)=0$, or
$\lim_{x\to x_0}f(x)=\pm\infty$ and $\lim_{x\to x_0}g(x)=\pm\infty$


Therefore it's traditional to summarize this lack of general theorems in these cases by saying that $0/0$ and $\infty/\infty$ are indeterminate forms.

There's nothing mysterious: we just know that, in order to compute (or show the existence of) a limit that appears to be in one of the indeterminate forms, we have to do more work than simply calculate a quotient. For instance apply l'Hôpital's theorem, or cleverly rewrite the function, or using other methods such as Taylor expansion.

Also the case when $\lim_{x\to x_0}f(x)\ne0$ and $\lim_{x\to x_0}g(x)=0$ is somewhat delicate, but not really “indeterminate”: the limit of the quotient either doesn't exist or is infinite ($\infty$ or $-\infty$).
Regarding the last point $f(x)=(x^2-9)/(x-3)$ you made an error; $f(x)$ can be rewritten as $f(x)=(x+3)(x-3)/(x-3)$, hence $f(x)=x+3$ that is the equation of a line defined in any $x$.