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.
For an extension $E/F$ and $\alpha\in E$, if $\alpha$ is algebraic over $F$, then the extension $F(\alpha)\subseteq E$ can also be described in a way which is intrinsic to $F$ (i.e. which involves no mention of $E$). 

To say that $\alpha$ is algebraic means there exists a non-zero (hence non-constant) polynomial $f\in F[X]$ with $f(\alpha)=0$. There is a unique $F$-algebra map $\varphi:F[X]\rightarrow E$ satisfying $X\mapsto(\alpha)$. The kernel is non-zero (because $f$ is in it) and proper (since this is a map of non-zero rings), so is of the form $(g)$ for a unique monic $g$, say of degree $d\geq 1$. One can characterize $g$ as the monic polynomial of minimal degree having $\alpha$ as a root. The homomorphism $\varphi$ gives rise to an isomorphism $F[X]/(g)\cong\mathrm{im}\varphi$. The image of $\varphi$ is $F[\alpha]\subseteq E$, the $F$-subalgebra of $E$ generated by $\alpha$, which can concretely be described as the set of polynomials in $\alpha$ with coefficients in $F$. This is a domain, being a subring of a field. One can also show that the elements $1+(g), X+(g),\ldots,X^{d-1}+(g)$ form a basis for $F[X]/(g)\cong F[\alpha]$ as an $F$-vector space. In particular, $F[\alpha]$ is of dimension $d$ as an $F$-vector space. From this it follows that $F[\alpha]$ is a field. Indeed, consider any non-zero element $\beta$ of $F[\alpha]$ and the $F$-endomorphism $x\mapsto\beta x:F[\alpha]\rightarrow F[\alpha]$. This is injective, hence surjective, since source and target have the same finite dimension, so there is $x$ with $x\beta=1$, and $\beta$ is a unit. Since, by definition, $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$, and $F[\alpha]$ is such a field, we must have $F[\alpha]=F(\alpha)$. 

In summary, we have constructed an isomorphism (of $F$-algebras) $F[X]/(g)\cong F(\alpha)$ (along the way we showed that $F(\alpha)$ consists of all polynomials in $\alpha$ with coefficients in $F$). Since $F[X]/(g)$ is a field, $(g)$ must be a maximal ideal, and $g$ is irreducible. It follows that $g$ is the unique monic irreducible polynomial in $F[X]$ having $\alpha$ as a root. In this way you get a description of $F(\alpha)$ as $F[X]/(g)$ which doesn't directly involve $E$. 

Conversely, if $g\in F[X]$ is a monic irreducible polynomial, then $(g)$ is maximal, so $E:=F[X]/(g)$ is a field containing $F$, and if $\alpha:=X+(g)$, then $\alpha$ is a root of $g$ in $E$, and $E=F(\alpha)$.

Note that this is only valid for $\alpha$ algebraic over $F$. As pointed out in one of the other answers, if $\alpha$ is transcendental over $F$, then the map $\varphi:F[X]\rightarrow E$ given by $X\mapsto\alpha$ is injective (this is exactly what it means for $\alpha$ to be transcendental over $F$), so one has $F[X]\cong F[\alpha]$ as $F$-algebras. This isomorphism of domains then extends uniquely to an $F$-algebra isomorphism of fraction fields $F(X)\cong F(\alpha)$.