Maybe some trivial examples could show you the importance of orientation. Try to compute the fluid flow through the unit square
$$
S = \left\{(x,y,z) \ | \ 0\leq x,y \leq 1 , z=0 \right\}
$$
of some fluid moving upwards at constant speed of $1$m/s.
$$
F(x,y,z) = (0,0,1) \ .
$$
Then you parametrize your surface, for instance with
$$
\varphi (x,y) = (x,y,0) , \quad (x,y) \in D = \left\{ (x,y) \ | \ 0\leq x,y \leq 1 \right\}
$$
The normal field to this parametrization is $\varphi_x \times \varphi_y = (0,0,1)$ and we compute the fluid flow:
$$
\int_\varphi \langle F, dS\rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,1) \rangle dxdy = 1 \ .
$$
But why using that particular parametrization? Why not trying with this one?
$$
\psi (u,v) = (v,u,0) , \quad (u,v) \in D = \left\{ (u,v) \ | \ 0\leq x,y \leq 1 \right\} \ .
$$
We compute the normal field again: $\psi_u \times \psi_v = (0,1,0)\times (1,0,0) = (0,0,-1)$. And the fluid flow:
$$
\int_\psi \langle F, dS \rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,-1) \rangle dudv = -1 \ .
$$
Astonishing, isn't it? -The fluid flow seems to depend on something secondary: the coordinates you use on the surface $S$. This could be even worse, since some surface integrals are used to compute the area of the surface: if the area of a surface depended on the coordinates you put in it, this would be a nonsense.
Fortunately, this isn't true: you can prove that surface integrals do NOT depend on the coordinates you put on the surface EXCEPT (in the case of fluid flows) on the orientation of the coordinates -parametrizations. There are two kinds of orientations and before computing a fluid flow you must choose one of them. For instance, in our examples, if we have said that the fluid is going upwards we have to choose $(0,0,1)$ as the orientation of our surface (the $\varphi$ parametrization). Otherwise, the result we get -with the $\psi$ parametrization- makes no sense: in one second $-1$ litre has gone trhough the surface!?