Well, yes, everyone (or, at least, everyone who can use the public key) knows the hash function H and G; so we can assume that an adversary knows them as well.
If YES: How does it help the security, if he just can decode
the padding and read the message?
Well, he can't decode the padding; the ciphertext has been encrypted using RSA, and he doesn't have the private key; so he can't get at the padded version.
Here's how OAEP works; you take the plaintext message $m$ to send, you pick a random value $r$, and compute the function:
$Padded = OAEP(m, r)$
(with H and G as parts of OAEP). Then, once we have that, we encrypt it using the base RSA public operation (using the public key):
$Ciphertext = RSA(PublicKey, Padded)$
The $Ciphertext$ is what's actually sent; because it is encrypted using RSA, the attacker cannot recover $Padded$.
So, if the OAEP operation doesn't prevent an attacker from decrypted the ciphertext in this straightforward manner, why is it there at all?
Well, raw RSA has some nastly properties (called an homeomorphic property), namely:
$RSA(k, A) \times RSA(k, B) = RSA(k, A\times B)$
(where $\times$ is implicitly modulo the public key modulus).
Because of this property, using raw RSA to directly encrypt plaintexts is almost always the wrong thing to do; there are clever ways to use this homeomorphic property to recover plaintext.
What the OAEP function does is try to mimic a random function between plaintexts and padded versions; because we present the raw RSA operation with effectively random texts, these homeomorphic properties are no concern.