There is a sense where you can start with just a scalar product.
If you assume that $v^2=vv=v\cdot v=|v|^2$ holds for any vector then that is a scalar product for a vector with itself.
From that you can get the scalar product of any two vectors $v$ and $w$, $v \cdot w= \frac{1}{2}\left((v+w)^2-v^2-w^2\right)=\frac{1}{2}\left(vw+wv\right)$ where for the last equality we assume the same distributive laws as for matrices.
So the simplest assumption of having a scalar product of any vector with itself can easily give us a scalar product of any two vectors (we can use the scalar products of $v+w, v$ and $w$ with themselves).
But we can also have a general product and allow it to have the regular rules of matrix algebras (associative, distributive, multiplication, addition, scaling by scalars, etcetera). Then we note that for vectors $v$ and $w$ we have $vw=\frac{1}{2}\left(vw+wv\right)+\frac{1}{2}\left(vw-wv\right)$ or $vw=v\cdot w+\frac{1}{2}\left(vw-wv\right)$ and we can notice that the last term $\frac{1}{2}\left(vw-wv\right)$ is different than a scalar or a vector, specifically its square is negative. For vectors $v$ and $w$ we can denote $\frac{1}{2}\left(vw-wv\right)$ by $v \wedge w$.
These new objects, like $v\wedge w$ naturally represent the plane spanned by the vectors $v$ and $w$. They are useful in physics, though for historical reasons we often use the complete accident that in 3d there are vectors orthogonal to planes to represent $v \wedge w$ by the vector orthogonal to it and then call it the cross product.
When you have both products $v \cdot w$ and $v\wedge w$ you have the full information to get the product $vw=v \cdot w+v\wedge w$, and from that you can solve for $v$ or $w$ if you have the other one. For instance $v=vw\frac{w}{w^2}=\left(v \cdot w+v\wedge w\right)\frac{w}{w^2}$. So therefore if you know the scalar product and the wedge product and you know $w$ then you can find out what $v$ was. Neither of the products by themselves allow you that. Many vectors can give the same inner product, and many vectors can give the same wedge product, but knowing both can allow you know the full relative relationship between the two.
It all follows from just the scalar product, but you get the whole package if you want it.
tl;dr
The dot product (symmetric part of the one product) tells you how much they have in common. The other product (antisymmetric part of the one product) tells you how much they orthogonal, specially how much you have to rotate one line to get align it with the other, and if you don't live in just a plane it also tells you the plane in which you need to rotate to send one into the other. That directionality is something you don't get from just a scalar.
But you don't need two products. One product is enough as long as you do it the invertible way. And that way naturally creates numbers, lines, planes and even higher dimensional objects as they come up. Further details above.
As pointed out, the projection and component actually refers to the same thing. To solve a problem like this it useful to introduce a coordinate system, as you mentioned yourself you project onto the x-axis. As soon as you introduce a coordinate system you can talk about the $\textit{components}$ of some vector. E.g $\vec{F} = F_1 \cos(\theta)\hat{x} + F_1 \sin(\theta) \hat{y}$ and $\vec{S} = S_1 \hat{x}$ if the box in your example is constrained to move horizontally. Here the components of the vectors are the $\textit{projections}$ of the vectors onto the coordinate axis. With this construct you calculate the dot product as $W = \vec{F} \cdot\vec{S} = F_1\cos(\theta) S_1$.
However it is not needed to introduce a coordinate system and writing the vectors by their components and then applying the rules for dot product. The dot product is defined in a coordinate independent way as a projection. So your question is just a matter of terminology. A $\textit{component}$ of a vector along some axis is the $\textit{projection}$ of the vector along that axis and in this sense projection is the more fundamental thing.
Best Answer
Well, say that you are looking at a man lifting boxes. Each box weighs 10kg. At first you look at him standing, but since just looking at him made you tired, you decide to lie down. Now, from your horizontal position, the scene looks different, but is the man doing more, less, or the same amount of work per box?
The word “scalar” is often used to just mean “number”, but it actually has a technical meaning: a scalar is a quantity that does not change when the system is transformed following one of the symmetries of the theory.
In this case, I think you are talking about classical mechanics, and the transformations involved are rotations in 3D space. As you know, the dot product of two vectors is invariant under rotation, that is why it's called “the scalar product”. The cross product, also called the vector product, transforms as a vector.
Now there is a final bit of the puzzle. The magnitude of a vector is also a scalar. Rotating the vector does not change its length. So in fact we have two ways to obtain a scalar from two vectors $\vec F$ and $\vec s$: taking the scalar product $\vec F \cdot \vec s$ or the magnitude of the vector product $|\vec F \times \vec s|$. How do we chose one?
Well, we know that when the force and the displacement are aligned, the work is maximum, and when they are perpendicular the work is minimum. This allows us to chose one alternative.