I really like this question! I can't yet upvote, so I'll offer an answer instead. This is only a partial answer, as I don't fully understand this material myself.
Suppose that $f$ is a quadratic polynomial. Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a "quadratic-like map." Then we'll call $f$ renormalizable. (See Chapter 7 of McMullen's book "Complex dynamics and renormalization" for more precise definitions.) Now renormalization preserves the property of having a connected Julia set. Also the parameter space of "quadratic-like maps" is basically a copy of $\mathbb{C}$.
So, fix a quadratic polynomial $f$ and suppose that it renormalizes. Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$. This gives a map from a small region about $f$ to the space of quadratic-like maps. This gives a partial map from the small region to the Mandelbrot set and so explains the "local" self-similarity.
To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way. (I believe that there are issues as you approach the place where the baby is attached to the parent.) Thus renormalization explains why the baby Mandelbrot set appears.
Let $S_1$ be a circle of radius $100$ in the complex plane centered at the origin. Clearly everything outside of $S_1$ diverges to $\infty$ under iteration of $P_c$, so the filled Julia set lies entirely inside of $S_1$.
Now take the preimage $S_2$ of $S_1$ under $P_c$. This will be a smaller curve (close to a circle, with a radius of approximately $\sqrt{100} = 10$), which again must contain the entire filled Julia set. Iterating this process, we obtain a sequence $S_1,S_2,\ldots$ of closed curves, each of which contains the filled Julia set in its interior. In fact, since any point outside of the filled Julia set goes to $\infty$ under iteration of $P_c$, the intersection of the interiors of the curves $S_n$ is precisely the filled Julia set. (See this picture for an example. The curves separating the different shades of orange are the iterated preimages of some large circle.)
Unfortunately, this reasoning is not quite correct, because the preimage of a closed curve under $P_c$ is not always a single closed curve. Sometimes it is one closed curve, and sometimes it is two closed curves with disjoint interiors. If we repeatedly take the preimages of $S_1$, we may find that $S_n$ is a union of a very large number of closed curves! Note, however, that these curves still contain the filled Julia set entirely inside of them.
Now here is the key bit: the preimage of a curve will have one component if and only if $0$ lies in the the interior of the preimage. This is because $0$ is the critical point of the map $P_c$. Thus the preimage of a curve is either a single curve that surrounds $0$, or two curves, neither of which surrounds $0$. (In the latter case, the two curves are actually negatives of one another, i.e. symmetric across the origin.) Therefore, there are exactly two cases:
The point $0$ lies in the filled Julia set. In this case, each preimage $S_n$ must be a single curve, so the intersection of the interiors is connected.
The point $0$ lies outside the filled Julia set. In this case, some preimage $S_n$ does not encircle $0$, so it must have two components. Then each successive preimage will have twice as many curves as the last, and the resulting filled Julia set is homeomorphic to the Cantor set. (See this picture for an example. The curves separating the different shades of blue are the iterated preimages of some large circle. You ought to be able to see the first step at which the curve separates into two components.)
Best Answer
$c$ in the above formulas is the complex number for which you want to determine if it is or isn't in the Mandelbrot set. In images of the Mandelbrot set, the real part of $c$ is typically mapped to the $x$ axis and the imaginary part to the $y$ axis. In other words, each point of the image corresponds to a different value of $c$. Most pictures of the Mandelbrot set additionally use colours to indicate how quickly the iteration diverges for points not in the Mandelbrot set. That is, the colourful images you usually see for the Mandelbrot set are in some sense images for the complement of the Mandelbrot set: That's where all the structure lies. The Mandelbrot set itself, on the other hand, usually is shown in an uniform colour (most often black, sometimes white, rarely other colours).