To measure flow through a surface, we need to know which way counts as "through": which side is the front. Choosing a side means choosing a unit normal at every point, in a way that does not flip suddenly. That choice is an orientation. It plays the same role for surfaces that the direction of travel plays for curves. You need parametric surfaces and their normal vectors.
Key idea
An orientation of a surface $S$ is a continuous choice of unit normal $\mathbf{n}$ at each point. (Some books write $\hat{\mathbf{n}}$.) A parametrization gives the orientation
$$\mathbf{n} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{|\mathbf{r}_u \times \mathbf{r}_v|},$$
or its opposite, if you swap $u$ and $v$. For a closed surface (one that encloses a solid), the positive orientation is outward. For a graph $z = f(x, y)$, the upward normal is a positive multiple of $\langle -f_x, -f_y, 1 \rangle$.
Why it works
Graphs: with $\mathbf{r} = \langle x, y, f(x, y) \rangle$, $\mathbf{r}_x = \langle 1, 0, f_x \rangle$ and $\mathbf{r}_y = \langle 0, 1, f_y \rangle$, so $\mathbf{r}_x \times \mathbf{r}_y = \langle -f_x, -f_y, 1 \rangle$, which has positive $z$-component: it points up.
Spheres: for $$\mathbf{r}(\varphi, \theta) = a\langle \sin\varphi\cos\theta, \sin\varphi\sin\theta, \cos\varphi \rangle,$$ a computation gives $\mathbf{r}_\varphi \times \mathbf{r}_\theta = a\sin\varphi\,\mathbf{r}$, a positive multiple of the position vector: outward.
Not every surface has two sides. Slide a normal vector once around the center line of a Möbius strip and it comes back pointing the other way. No continuous choice exists, so the Möbius strip is not orientable, and flux through it is not defined.
Spheres: for $$\mathbf{r}(\varphi, \theta) = a\langle \sin\varphi\cos\theta, \sin\varphi\sin\theta, \cos\varphi \rangle,$$ a computation gives $\mathbf{r}_\varphi \times \mathbf{r}_\theta = a\sin\varphi\,\mathbf{r}$, a positive multiple of the position vector: outward.
Not every surface has two sides. Slide a normal vector once around the center line of a Möbius strip and it comes back pointing the other way. No continuous choice exists, so the Möbius strip is not orientable, and flux through it is not defined.
Worked examples
Worked example: Upward on a graph
Find the upward unit normal to $z = x^2 + y^2$ at $(1, 1, 2)$.
- $f_x = 2x = 2$ and $f_y = 2y = 2$, so $\langle -f_x, -f_y, 1 \rangle = \langle -2, -2, 1 \rangle$.
- Its length is ${}3$, so $\mathbf{n} = \frac{1}{3}\langle -2, -2, 1 \rangle$. (Downward would be the negative.)
Answer: $\frac{1}{3}\langle -2, -2, 1 \rangle$
Worked example: Parameter order matters
For $\mathbf{r}(u, v) = \langle u, v, 3 - u - 2v \rangle$ and $\mathbf{s}(u, v) = \langle v, u, 3 - v - 2u \rangle$ (the same plane), which cross product $\mathbf{r}_u \times \mathbf{r}_v$, $\mathbf{s}_u \times \mathbf{s}_v$ points up?
- $$\mathbf{r}_u \times \mathbf{r}_v = \langle 1, 0, -1 \rangle \times \langle 0, 1, -2 \rangle = \langle 1, 2, 1 \rangle\text{:}$$ up.
- $$\mathbf{s}_u \times \mathbf{s}_v = \langle 0, 1, -2 \rangle \times \langle 1, 0, -1 \rangle = \langle -1, -2, -1 \rangle\text{:}$$ down.
- Swapping the roles of the parameters flips the normal.
Answer: $\mathbf{r}$ points up, $\mathbf{s}$ points down
Worked example: A cylinder
Does $\mathbf{r}_\theta \times \mathbf{r}_z$ point outward on $\mathbf{r}(\theta, z) = \langle a\cos\theta, a\sin\theta, z \rangle$?
- $\mathbf{r}_\theta = \langle -a\sin\theta, a\cos\theta, 0 \rangle$, $\mathbf{r}_z = \langle 0, 0, 1 \rangle$.
- $\mathbf{r}_\theta \times \mathbf{r}_z = \langle a\cos\theta, a\sin\theta, 0 \rangle$, pointing straight away from the axis: outward.
Answer: yes
Common mistake
Letting the parametrization choose. The formula $\mathbf{r}_u \times \mathbf{r}_v$ gives a normal, but it may point the wrong way for the problem: inward on a closed surface, or down when the problem says up. Check one convenient point. If the cross product points the wrong way, use $\mathbf{r}_v \times \mathbf{r}_u$, which flips the sign of every flux integral.