Question: The way I learned it was given a field extension $F \subset E$, and an element $\alpha \in E$

$$F(\alpha) := \{p(\alpha)/q(\alpha) : p(x), q(x) \in F[x] ,q(\alpha) \not = 0\} $$

Is there an easier way to think about the field $F(\alpha)$

Answer: 
The definition you've given yields the form of an arbitrary element of the extension. However, it is not the intuitive definition. The intuitive definition is simple: $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$.

In fact, your definition can also be somewhat simplified if $\alpha$ is algebraic over $F$. In this case $F(\alpha)=F[\alpha]$, that is, the smallest ring containing $F$ and $\alpha$, so $F(\alpha)=F[\alpha]=\lbrace p(\alpha)\vert p\in F[x]\rbrace$. Furthermore, algebraic elements have finite degree, so we only need to consider $p$ of degree smaller than the degree of $\alpha$ over $F$.

If $\alpha$ is transcendental over $F$, then $F(\alpha)$ is actually isomorphic to the field of rational functions over $F$, $F(x)$, in which case it is pretty much what you've written, but of course there's no restriction on $q$ in this case.
Ok so the intuition is that you start with the field $F$ and throw into the pot this new "thing" $\alpha$ to make a field $F(\alpha)$, defined to be the smallest field containing $F$ and $\alpha$.

Let's try an example and discover what the field $\mathbb{Q}(\sqrt{2})$ looks like. Ok so this field has to contain all rational numbers. It also contains $\sqrt{2}$ and all rational multiples of this. Thus in fact $\mathbb{Q}(\sqrt{2})$ must contain all numbers of the form $a+b\sqrt{2}$ for rational $a,b$. It must also contain $\sqrt{2}^2, \sqrt{2}^3, ... $ and all rational multiples of these. But these powers of $\sqrt{2}$ are redundant in this counting (they can all be written in terms of numbers of the form $a+b\sqrt{2}$ and so are already accounted for).

Now $\{a+b\sqrt{2}\,|\,a,b\in\mathbb{Q}\}$ is actually a field itself (easily checked), so by minimality it must be equal to $\mathbb{Q}(\sqrt{2})$.

Now once you try to work out a few more of these you start to realise that the above process worked smoothly purely because of the fact that $\sqrt{2}$ is algebraic over $\mathbb{Q}$. 

In general if $\alpha$ is algebraic over your initial field $F$ then this "redundancy" occurs...and that you only have to use the powers of $\alpha$ upto $d-1$, where $d$ is the degree of the "minimal polynomial" of $\alpha$ over $F$ (this being the smallest polynomial with $\alpha$ as a root over $F$). 

We had to stop at the first power of $\sqrt{2}$ above because $\sqrt{2}$ satisfies a quadratic polynomial $x^2 - 2$ over $\mathbb{Q}$ and it satisfies no polynomial of smaller degree. 

When $\alpha$ is transcendental over $F$ then all powers of $\alpha$ (positive and negative) and all sums, multiples, quotients of these must be thrown into the pot.