The three points you list $(0,0,0), (0,1,1)$ and $(0,2,0)$ do not form an ellipse. They form a triangle in the $yz$ plane as shown in the figure below, where the B and C in that figure are the projections of the $B$ and $C$ in the question. You can find the scalar area of this triangle and then attach an appropriate unit normal (which will be $\pm\hat{\mathbf{i}}$ ) and this is the $x$ component of $\mathbf{S}$. You can find the $y$ and $z$ components of $\mathbf{S}$ similarly. Finally $\mathbf{S}=({S}_x,{S}_y,{S}_z)$. Note, not all projections will necessarily be triangles, but they will all be simple planar figures whose scalar area can easily be determined.
This really boils down to the long-standing question of "what is the best intuitive understanding of the dot product" which, admittedly, is no dobut a mystery to many as it very often comes up.
What I would say with regard to that question - and which thus also answers the one in your title - is that the dot product is a "mathematically nicefied" version of the projection operation: namely, the intuition is that $\mathbf{a} \cdot \hat{\mathbf{u}}$, for a unit vector $\hat{\mathbf{u}}$, is indeed that it is the component of $\mathbf{a}$ in the direction of $\hat{\mathbf{u}}$, and we could start the construction of the dot product from there.
The question, of course, arises as to what to do if $\mathbf{u}$ is not a unit vector. One option would be just to say that it doesn't matter: just give the projection (component) in the direction anyways. However, that leads to some problems. For one, let's consider commutativity. Ideally, we'd like it if our product were commutative:
$$\mathbf{a} \cdot \hat{\mathbf{u}} = \hat{\mathbf{u}} \cdot \mathbf{a}$$
however, if we are going down the line of description I just gave and just say it gives the projection to the second operand as a pure directional indicator, then we have an asymmetric, non-commutative operation.
Yet fortunately, it turns out that, very nicely, these problems go away if we decide that
$$\mathbf{a} \cdot \mathbf{u}$$
when $\mathbf{u}$ is not a unit vector, is to be the projection scaled by the length of the vector of projection, here, $\mathbf{u}$. Given that the projection already scales by the length of the vector being projected, effectively, this causes both vectors to be treated "on equal footing", so to speak, and the operation now commutes. Moreover, it becomes even bilinear, which is even better, and thus makes it a tensor (perhaps even the canonical example!), which allows all sorts of generalizations in more sophisticated mathematical usage, including the ability to define Riemannian manifolds and talk about curved, non-Euclidean spaces, and also the ability to provide infinite-dimensional spaces with useful topological structures permitting things like infinite linear combinations, as we get when we go down the inner product space sequence.
In short, the dot product represents projection onto a unit vector because at that point, the scaling coefficient we introduce to make it mathematically "nice" and "even-handed" equals $1$ when the vector of projection is such.
Best Answer
Remember that the direction of a vector $b$ is given by $\hat{b}=\frac{b}{||b||}$. You can decompose $\hat{b}$ into a component that is parallel to $a$ and one that is orthogonal to it. Since the cosine of 90º is zero, this operation will "select" the component that is parallel to $a$. This gives you a number, which represents (plus or minus) the length of the orthogonal projection of $\hat{b}$ onto $a$. This is called the scalar projection. The vector projection is obtained by multiplying the vector$\hat{b}=\frac{b}{||b||}$ by the scalar projection. This results in $$(\hat{b}\cdot a)\hat{b}= \frac{b\cdot a}{\sqrt{b\cdot b}}\hat{b}= \frac{b\cdot a}{b\cdot b}b.$$
Note that this is a vector, so it shouldn't equal any distance at all. If you take a vector that is perpendicular to a line, represent it in a way that it starts at a point $a$ in that line and ends at a point $b$ above it, then you can use the absolute value of the scalar projection to prove this distance formula you mentioned.