Intuition behind orientation of a surface Let $S\subseteq \mathbb{R}^3$ be a smooth surface. 
Suppose $\phi: U\rightarrow S$ is a (local) parametrization for $S$, where $U \subseteq \mathbb{R}^2$ is an open set. Then $\phi(U)$ has a standard orientation, that is, given $p\in\phi(U)$ ($p=\phi(x_0)$, $x_0 \in U$) the orientation of $T_pS$ is defined to be the orientation given by the basis $\{d\phi(x_0)(e_1),d\phi(x_0)(e_2)\}$ ($\{e_1,e_2\}$ being the standard basis of $\mathbb{R}^2$).

I would like to understand how this definition implies that if $p, q \in S$ are close, than the orientations for $T_pS$ and $T_qS$ are "similar", without referring to normal vectors.

I guess the problem is how to formalise this "similarity". We're trying to extend the concept of orientation from vector spaces to a piece of $S$. However is $p\neq q \in S$ then $T_pS$ and $T_q S$ are different vector spaces, so it doesn't make much sense to compare the orientations of two respective basis. 
One idea I had is to "slide" $T_qS$ to $T_pS$ along the surface $S$ to be able to compare two respective basis. This argument will only work when $\phi(U)$ is connected, i.e. only when $U$ is connected (as $\phi$ is a homeomorphism onto $\phi(u)$), but this is not a big issue because if $\phi(U)$ is not connected than it doesn't make much sense (to me) to talk about a "global" orientation of disconnected pieces.
So I can suppose $U$ and $\phi(U)$ are connected. Since we're working in euclidean spaces then it means that $U$ and $\phi(U)$ are path connected, so there exists a path $\gamma$ in $U$ that lifts to a path from $p$ to $q$ in $S$. However this path need not be smooth, so it might not behave well for this purposes. So I cannot come up with a canonical way of "sliding", hence I'm stuck here, provided this is a good idea.
Do you know how to go on from here or an alternative explanation?
 A: Alternative explanation:
On an open set $U$ of an Euclidean space $\mathbb{R}^2$, what you earn is that you can identify the tangent spaces at each point since they are canonically isomorphic to $\mathbb{R}^2$. Then, an orientation on this open set is a choice of a family of basis of $\mathbb{R}^n$, say $(\mathcal{B}_x)_{x\in U}=((e_1(x),e_2(x)))_{x\in U}$, such that for all $x$ and $y$ in $U$, you have $\det_{\mathcal{B}_x}(\mathcal{B}_y)>0$. You can say that this orientation is smooth if the maps $e_1:U\to\mathbb{R}^n$ and $e_2:U\to\mathbb{R}^n$ are smooth. For example, $\mathcal{B}_x=(e_1,e_2)$ for all $x$ in $U$ is a smooth orientation. 
Now, once you get a parametrization $\varphi:U\to S$, it gives you a smooth identification between $U$ and $\varphi(U)$. For example, it sends a smooth orientation $(e_1(x),e_2(x))_{x\in U}$ on a smooth family of basis of tangent spaces $T_{\varphi(x)}S$, by the formula you gave: $\mathcal{B}_{\varphi(x)}=(d\varphi_x(e_1(x)),d\varphi_x(e_2(x)))$. 
And this is the actual way to define an orientation on $S$: it is a family of basis of $T_pS$, say $(\mathcal{B}_p)_{p\in S}=(f_1(p),f_2(p))_{p\in S}$, such that for all parametrizations $\varphi:U\to S$, the unique families of vectors of $\mathbb{R}^2$ $(e_1(x),e_2(x))_{x\in U}$ checking $(d\varphi_x(e_1(x)),d\varphi_x(e_2(x)))=(f_1(\varphi(x)),f_2(\varphi(x)))$ are a smooth orientation on $U$. 
About the sliding idea: 
If you are only given a tangent vector of $T_pS$, and that you try to make it slide along a path in order to get a tangent vector of $T_qS$, it is a difficult task and you will need an additional piece of data, called a connection, which will allow you to define what you are looking for, namely parallel transport. 
