How can $n$-dimensional space be projected? How is spacial projection generalized? For example I can project three-dimensional space into two-dimensional space (like with a camera), I can project two dimensional space into a dotted line or something. I can't manage to fathom how $n$-dimensional space is projected into a form my mind can understand. How is this type of projection generalized for use with an ambiguously defined number of spacial dimensions?
 A: Let's start (and stop) at 4D, because even that is tricky.

A face of a 4D cube is a 3D cube and each 4D cube has 8 such faces. The picture above shows a 4D cube projected into a plane with the four principal, orthogonal axes shown. It takes a lot of staring to see the faces of a 4D cube. The pictures below try to illustrate how the two faces along each of the four axes are situated in 4D. 


I can't manage to fathom how n-dimensional space is projected into a form my mind can understand.

Nobody else can either, working in 4D is difficult enough 

How is this type of projection generalized for use with an ambiguously defined number of spacial dimensions?

It's not. We use math instead and totally give up on the "try to visualise"  idea. Your brain is hardwired for 3D, so it's impossible.
A: If you have a D-dimensional vector $\vec{d} \in \mathbb{R}^D$ you can project that vector into a (D-1)-dimensional hyperplane perpendicular to a unit-vector $\vec{n}$ in the following way:
The component of $\vec{d}$ along the direction of $\vec{n}$ is its projection
$$ d_n = \vec{n} \cdot \vec{d} $$
So the part of $\vec{d}$ pointing along $\vec{n}$ is
$$ \vec{d_n} = d_n \cdot \vec{n} $$
The part of $\vec{d}$ perpendicular to $\vec{n}$ is therefore:
$$ \vec{d}_\perp = \vec{d} - \vec{d_n} $$
since $\vec{n} \cdot \vec{d}_\perp =  \vec{n} \cdot \vec{d} - \vec{n} \cdot \vec{d_n} = d_n - d_n = 0$.
Thus we define the projection function $P_{\vec{n}}: \mathbb{R}^D \rightarrow H_\vec{n} \simeq \mathbb{R}^{D-1}$ by
$$ P_{\vec{n}}(\vec{x}) = \vec{x} - (\vec{n}\cdot\vec{x}) \cdot \vec{n} $$
which projects every vector into the hyperplane $H_\vec{n} \simeq \mathbb{R}^{D-1}$ perpendicular to $\vec{n}$.
For $D=3$ this would be a (rotated) plane, where $\vec{n}$ is the plane's normal vector.
For $D=4$ this would be a (4-dimensional rotated) 3-dimensional space, etc.
