What does it mean to parameterise a function and constraining it to a surface? In this video the author explains the concept of surface integrals, which I'm struggling with. 
In the first few minutes he explains that the complexity of calculating $ \iint_s f(x,y,z) dS $ arises from "constraining to the surface S". What does this mean?
From what I can gather $ f(x,y,z) $ is a 4D surface so I assume it's like a projection, but I'm not sure.
The author then goes on to say that a parameterisation of this function (what function? The integral?) is;
$$ \vec{r} = <x(u,v), y(u,v), z(u,v)>$$
Here is the screenshot I'm referring to:

What are $u$ and $v$? Are they axes? Where? And what's $r$? 
I'm more confused than I was when I started watching this video, so I'm thinking it's best to stop and just ask.
 A: There is indeed a surface in $\mathbb{R}^4$ defined by graphing $f$, but the surface they are referring to is the $S$ in $\int_S$ which is a two dimensional surface in three dimensional space. The integrand $f\left(x,y,z\right)$ is defined over some subset of $\mathbb{R}^3$ but the integral is over just $S$, a two dimensional submanifold of $\mathbb{R}^3$. 
The $u,v$ are the coordinates of a parametrisation. This means that every pair $\left(u,v\right)$ corresponds to a unique point on $S$ so that $S$ is 'parametrised' by $u,v$. It should make sense that this can be done as $S$ is two dimensional. Picture laying out grid lines over $S$ which are allowed to be curved and using these to assign $u,v$ coordinates to every point of $S$. 
The $\bf{r}$ is a shorthand for $\left(x,y,z\right)$. 
As a side note, there are different kinds of surface integral where the integrand is a vector field, or a one-form. 
A: Whilst it is true that $f=f(x,y,z)$ is indeed a $4$-D object if you try to look at it that way. But I think it may be easier to simply think of $f(x,y,z)$ as a scalar field.
i.e. a number $f$ is associated with each coordinate $(x,y,z)$.
On the other hand, $S$ is a surface in $3$-D, i.e. a collection of points $(x,y,z)$. So, intuitively, the integral
$$\iint_S f(x,y,z) \, dS$$
is just asking you to "sum up" all values of $f$, for the given set of $(x,y,z) \in S$.
Constraining to the surface $S$ just means that you have to make sure that the points $(x,y,z)$ you are integrating over do indeed belong to $S$.
We must hence find a way to represent exactly the points in $S$.
For example, if you were asked to integrate over the $1$-D set $S=[0,5]$ for the function $f(x)$, then you would simply do
$$\int_0^5 f(u) \, du$$
where I have deliberately used $u$ instead of $x$ to make the distinction.
What you have actually done in the process is you have parameterised the set $[0,5]$ using the parameter $u$. You let $x(u)=u$, and said that $u$ is allowed to vary from $0$ to $5$ so that it covers all the points of $S$ exactly.
Equivalently, you could have decided to parameterise $x(u) = 5u$, in which case $u$ can only vary from $0$ to $1$ this time, because $x(u=0)=0$ and $x(u=1)=5$, so that you cover all the points of $S$ exactly.
As you may have noticed, this is basically integration by substitution. Indeed, if you choose to parameterise with $x(u)=5u$, then the integral becomes
$$\int_0^1 f(5u) \cdot 5du$$
because $dx = 5du$.
Now, we go back to the $3$-D case. It is much easier to understand with an example, say
$$S = \{(x,y,z) \in \Bbb R^3: x^2+y^2 \leq 1 , z=0\}$$
Again, we must find a way to parameterise (i.e. represent) all the points in $S$ exactly. As a general rule, if $S$ is an $n$-dimensional object, then you must parameterise $S$ with $n$ parameters.
In the $S=[0,5]$ case previously, we used a single parameter $u$ because $S$ is $1$-D. This time, $S$ is $2$-D, so we use two parameters, $u$ and $v$ for example.
Consider the representation
$$x(u,v) = u\cos v \qquad y(u,v) = u\sin v \qquad z(u,v) = 0$$
where $u \in [0,1]$, $v \in [0,2\pi)$. i.e.
$$\mathbf r(u,v) = \begin{pmatrix} u\cos v \\ u\sin v \\ 0 \end{pmatrix}$$
When parameterising, it is important to specify the range of the parameters. You can check for yourself that as $u$ varies  over $[0,1]$ and as $v$ varies over $[0,2\pi)$, all the points in $S$ are covered, no more and no less.
In the same way that we had $dx = 5du$ in the $1$-D case due to the parameterisation $x=5u$, we must to the same in the $2$-D case. As shown in the picture you attached, the formula is
$$dS = \bigg|\frac{\partial \mathbf r}{\partial u} \times \frac{\partial \mathbf r}{\partial v} \bigg| \, dudv$$
In this case, we can have
$$dS = \bigg|\frac{\partial \mathbf r}{\partial u} \times \frac{\partial \mathbf r}{\partial v} \bigg| \, dudv = \bigg|\begin{pmatrix} \cos v \\ \sin v \\ 0 \end{pmatrix} \times \begin{pmatrix} -u\sin v \\ u\cos v \\ 0 \end{pmatrix} \bigg| = |u|=u$$
Finally, the required integral becomes 
$$\iint_S f(x,y,z) \, dS = \int_{v=0}^{v=2\pi} \int_{u=0}^{u=1} f\big(u\cos (v), u\sin (v), 0\big) \cdot ududv$$
