Why does DSS have such formulas? I've learnt the formulas needed for the DSA (http://en.wikipedia.org/wiki/Digital_Signature_Algorithm).

But most importantly, I don't quite understand what represents those formulas and why they are the way they are. For example, why if v = r then the signature is verified, why do r and s have such formulas etc.? RSA digital signature is so much simpler.
The fundamental difference between RSA-groups and prime order groups is that in RSA groups the multiplicative order of the group is unknown (without knowledge of the factorization). This allows much easier constructions for unforgeable signatures (although hashing and padding are required to ensure existential unforgeability).

With knowledge of the group order, a forger has more possibilities to calculate "useful" group elements. This results in the DSA formulas (and the related ElGamals signatures) to be more complex.

Let's have a look at what these formulas actually calculate, and for simplification I'll leave out all the modulo operations (they are applied in the correct way). The goal is to get expressions which only depend on the hash $H$, the signing key $x$, and the (random) blinding value $k$.

The signature is:
$$r=g^k$$
$$s=k^{-1}(H+xr) = k^{-1}(H+xg^k)$$

Let's look at the verification algorithm backwards:
$$v=g^{u_1}y^{u_2} = g^{Hs^{-1}}g^{x \cdot r s^{-1}}$$
$$=g^{s^{-1} (H+xr)}$$

Here is the interresting part: A valid signature will solve like this:
$$= g^{(k^{-1}(H+xg^k))^{-1} (H+xg^k)}= g^k = r$$

For a forger, this is differently: He can choose $r$ and $s$ arbitrarily, but he does not know $x$ but only $y=g^x$. To forge a valid signature, it has to fulfill this equation:
$$g^{s^{-1}H}y^{s^{-1}r} = r$$
However, we can not solve this equation without calculating some discrete logarithms.