How RSA-OAEP decrypt a Symmetric Key message? In an RSA-OAEP encrypted symmetric Key (AES for example) message. The message is then decrypted by the other party when R is a Random Number That other party doesn't know about. so How it can retrieve back the key when it's XORed with A Hash (hash1) That it doesn't know. I don't think that output of Hash2 helps with anything. Right?
OAEP is Optimal asymmetric encryption padding for RSA and developed by M. Bellare, P. Rogaway, in 1995 and standardized in PKCS#1 v2 and RFC 2437.

Optimal Asymmetric Encryption -- How to encrypt with RSA. Extended abstract in Advances in Cryptology

Your image hiding some internals of the OAEP. Here a better one;

The MFG is Mask Generation Function is expected to be a random oracle. The MFGs are similar to a cryptographic hash function except that while a standard hash function's output is a fixed size, the MGF supports output of a variable-length.
If OAEP is developed after the XOF (Extendable Output Functions) then the proof will be much easier.
Back to your question

How it can retrieve back the key when it's XORed with A Hash (hash1) That it doesn't know. I don't think that the output of Hash2 helps with anything. Right?

We can formalize above as;
\begin{align}
T &amp;= lhash \mathbin\| PS \| \texttt{01} \mathbin\| Message\\
maskedDB &amp;= MFG1(seed) \oplus T \\
maskedSeed &amp;= MFG2(maskedDB) \oplus seed\\
\end{align}
In the PKCS#1 standard, the same MFG is used as the random oracles. I've made a distinction by numbering. $MFG1$ is taken the $seed$ as input.
Now, you got a message that is sent to you by RSA-OAEP. You get the $maskedSeed$ and $maskedDB$
The $seed$ can be calculated by
$$seed = maskedSeed \oplus maskedDB$$ and now we know the $seed$.
Now the $T$ can be computed by
$$ T = MFG1(seed) \oplus maskedDB $$
Now get the encoded $message$ block and check it.
$$T = lhash \mathbin\| PS \| \texttt{01} \mathbin\| Message$$