What exactly does it mean to take something modulo an equivalence relation? For instance, the complex projective space is defined as $\mathbb{C}{\mathbb{P}^n} = \left( {{\mathbb{C}^{n + 1}}\backslash \left\{ 0 \right\}} \right)/ \sim $
Where the equivalence relation is defined for $z,w \in {\mathbb{C}^{n + 1}}$ such that $z \sim w\;\;\;\;\; \Leftrightarrow \;\;\;\;z = \lambda w\;\;\;for\;\;\;\lambda  \in {\mathbb{C}^ * }$
Is there an intuitive way to interpret the effect of this geometrically? 
 A: The points of the new space are the equivalence classes of points of the old space. Here, the new points are the (punctured) lines through the origin. To have "really" points again, you might want to pick a point from each equivalence class - but that causes difficulties in the end as there is often no natural and consistent way to do so ...
A: In set theory, it just means you take the set of equivalence classes. If the set is $X$ and the equivalence is $\sim$ we call the set of equivalence classes $X/\sim$.
Outside set theory, like in algebra and topology, you have to give the set of equivalence classes structure, in this case, a topology.
If $\sim$ is an equivalence class on a topological space $X$, consider the natural map $p:X\to X/\sim$, and we define the topology on $X/\sim$ so that $U\subseteq X/\sim$ if $p^{-1}(U)$ is open in $X$.
This has an important "universal" property:

If $f:X\to Y$ is a continuous function of topological spaces such that $x_1\sim x_2$ implies $f(x_1)=f(x_2)$, then $f$ factors through $p$ - there is a unique continuous function $g: X/\sim\to Y$, such that $f(x)=g(p(x))$ for all $x\in X$.

A: One thing that the answers given haven't addressed: Yes, often you can understand this intuitively. Particularly in geometry and topology, the equivalence relation will only equate a finite number of points, and you can think of it as gluing these points together. 
For example, consider the square $[-1,1]\times[-1,1]$ modulo the following equivalence relations. In each of the relations if $|x| < 1$ and $|y| < 1$ then $(x, y) \sim (u, v)$ if and only if $(x, y) = (u, v)$. Only on the boundaries are points equivalent to other points:


*

*$(-1, y) \sim (1,y)$ - we are gluing the left edge to the right edge. Think of the square as a piece of paper we pick up the left and right edges and pull them together. The equivalence relation combines each point on the left edge with a point on the right edge, "gluing" them together to form a new point. The result is a cylinder.

*$(-1, y) \sim (1, -y)$ - the same as above, except this time we twist our paper over as we bring left side to right (with an actual square of paper you can't do this, but with "topologist's paper", it can bend and stretch all you like, as long as you don't tear it). The equivalence relation glues the reversed right edge to the left edge to form a Möbius strip.

*$(-1, y) \sim (1, y)$ and $(x, -1) \sim (x, 1)$. This forms the cylinder as before, but also glues the top circle of the cylinder to the bottom circle to form a torus.

*$(-1, y) \sim (1, y)$ and $(x, -1) \sim (-x, 1)$ - this reverses the top circle before gluing it to the bottom. This maneuver is impossible in $\Bbb R^3$ without self-intersection, but is possible in $\Bbb R^4$. The result is the Klein Bottle.

*$(-1, y) \sim (1, -y)$ and $(x, -1) \sim (-x, 1)$ - Reversing both edges before gluing gives the projective plane.


You can often still picture what is going on even when infinite collections of points are being merged into a single point. For example, what do you get when you make all the boundary points of the square equivalent to each other? Well, think of the square as a sheet of cloth that you pick up by the four corners and bring them all together. Then you gather up the sides and bunch the whole boundary up together "squeezing" it into a single point. It doesn't take too much to see that topologically, this is a sphere.
In your own example, but done in $\Bbb R^n$ instead of $\Bbb C^n$, we can modify the equivalence relation to $ x \sim y$ if there exists $\lambda > 0$ such that $x = \lambda y$. In this case, you are collapsing each ray eminating out of the origin down to a single point. We can see the effect of this by simply noting that each ray intersects the unit hypersphere in exactly one point, so intuitively we are collapsing the entire ray down to the point where it intersects the sphere, thus the result is a sphere.
The projective relation where $x \sim y$ if $x = \lambda y$ for any $\lambda \ne 0$ collapses the entire line (except for $0$) to a single point. Since we already had the line being equivalent to the two points where it intersects the sphere with the previous relation, this equivalence can be thought of as going one step further and gluing the antipodal points on the sphere together as well. Thus you can think of the real projective plane as being a collection of antipodal pairs of points on a sphere being treated as if they were single points.
In $\Bbb C^n$ the picture is much the same, except now it is entire planes being collapsed to a point instead of lines. As there are no good images of this in $\Bbb R^3$, we can't really visualize it, but we can still consider it as being similar to the Real version.
