Before giving a comparison/contrast type answer, let's first examine what the two theorems say intuitively.
Stokes' Theorem says that if $\mathbf{F}(x,y,z)$ is a vector field on a 2-dimensional surface $S$ (which lies in 3-dimensional space), then $$\iint_S \text{curl }\mathbf{F}\cdot d\mathbf{S} = \oint_{\partial S} \mathbf{F}\cdot d\mathbf{r},$$
where $\partial S$ is the boundary curve of the surface $S$.
The left-hand side of the equation can be interpreted as the total amount of (infinitesimal) rotation that $\mathbf{F}$ impacts upon the surface $S$. The right-hand side of the equation can be interpreted as the total amount of "spinning" that $\mathbf{F}$ affects along the boundary curve $\partial S$. Stokes' Theorem then tells us that these two seemingly different measures of "spin" are in fact the same!
It is remarkable also because solely from knowing how $\mathbf{F}$ affects the boundary curve $\partial{S}$, we can deduce how $\text{curl }\mathbf{F}$ affects the entire surface!
The Divergence Theorem says that if $\mathbf{F}(x,y,z)$ is a vector field on a 3-dimensional solid region $E$ (which lies in 3-dimensional space), then $$\iiint_E \text{div }\mathbf{F}\,dV = \iint_{\partial E} \mathbf{F}\cdot\mathbf{N}\,dS,$$ where $\partial E$ is the boundary surface of the solid region $E$, and $\mathbf{N}$ is an outward-pointing normal vector field on $E$.
If we think of $\mathbf{F}$ as being some sort of fluid, then the left-hand side measures how much of the fluid is outward-flowing (like a source) or inward-flowing (like a sink). That is, the left-hand side measures the total amount of (infinitesimal) divergence (outwardness/inwardness) of $\mathbf{F}$ throughout the entire solid $E$.
On the other hand, the right-hand side tells us how much of $\mathbf{F}$ is "passing through" the boundary surface $\partial E$. In other words, it is the flux of $\mathbf{F}$ across $\partial E$.
So, the Divergence Theorem tells us that these two different measures of the "outwardness" of $\mathbf{F}$ (the sources/sinks inside the solid vs the flux through the boundary) are in fact the same! To quote Wikipedia: "The sum of all sources minus the sum of all sinks gives the net flow out of a region."
And again, we have a situation where the behavior of $\mathbf{F}$ on the boundary gives us insight into how $\mathbf{F}$ acts on the entire region!
Similarities: Both Stokes' Theorem and the Divergence Theorem relate behavior of a vector field on a region to its behavior on the boundary of the region. As Zhen Lin pointed out in the comments, this similarity is due to the fact that both Stokes' Theorem and the Divergence Theorem are but special cases of a single, very powerful equation (known as the Generalized Stokes Theorem).
(The Generalized Stokes Theorem is somewhat advanced, and usually goes by the name Stokes' Theorem, whereas the Stokes' Theorem we've been talking about is often called the Kelvin-Stokes Theorem. This is why the Wikipedia page on "Stokes' Theorem" may seem rather advanced -- it is primarily about the Generalized theorem.)
Differences: Stokes' Theorem talks about "rotation" along a surface which has a boundary curve. The Divergence Theorem talks about "sources and sinks" inside a solid that has a boundary surface.
So, in addition to being about different types of quantities ("rotation" vs "divergence"), you should note that the two theorems apply to completely different types of regions. That is, a surface which has a boundary curve (setting of Stokes' Theorem) cannot enclose a solid volume (setting of the Divergence Theorem), and conversely.
The unit normal vector $\bf N$ is by definition ${\bf T}'\over \Vert {\bf T}'\Vert$.
Since $\bf T$ has constant norm one, it is orthogonal to $\bf T'$:
$${\bf T}\cdot {\bf T}'=0$$
(if $\bf X$ has constant norm, then $0={d\over dt}({\bf X}\cdot{\bf X })=
{\bf X}'\cdot{\bf X }+{\bf X}\cdot{\bf X }' =2{\bf X}' \cdot{\bf X} $)$^\dagger$.
This implies ${\bf T}\cdot {{\bf T}'\over \Vert {\bf T}'\Vert}=0$; thus $\bf N$ is normal to $\bf T$.
$^\dagger$ This is the real "reason" why $\bf N$ is orthogonal to $\bf T$. To repeat myself: if ${\bf X} $ has constant norm, then ${\bf X} (t)$ is orthogonal to ${\bf X}'(t)$. Though the product rule for differentiating a dot product is a perfectly fine way to see this, I prefer the following "proof":
Suppose ${\bf X}(t)$ has constant norm $a$ and consider it to be a position vector. Then as $t$ varies, ${\bf X}(t)$ describes a path on the surface of a sphere of radius $a$. We know that ${\bf X'}(t_0)$ is tangent to the path traced out by ${\bf X}(t)$ at the point ${\bf X}(t_0)$; so, ${\bf X'}(t_0)$ is tangent to the aforementioned sphere and, hence, orthogonal to ${\bf X}(t_0)$.
Best Answer
$(1)$ since the cube is bottomless, the square at the bottom of the cube is the boundary. This square lies in the plane $z=-1$, but it is not the whole plane.
$(2)$ firstly, your green and purple arrows are going in the same direction, so lets just imagine the green one is pointing the other way.
Both $n=k$ and $n=-k$ are normal to $C$, you can take either as long as the person is walking around the surface in the right way (discussed below).
The way to determine the correct direction (looking from below) is to imagine that the interior of the surface must always be on the person's left (as they walk around the boundary) which corresponds with the green direction.
I was explaining the direction choice for $n=−k$, so that it matched your picture, the example uses $n=k$, so the direction changes (from green to purple), this is so we can go around the boundary in the same directions for both surfaces $S_1$ and $S_2$.