If OAEP is introducing randomness to encryption how could be ciphertext decrypted? I was studying OAEP and see we introduce a random value $r$ when computing the padding. See the figure below from Wikipedia 

So if we are adding randomness to encryption, how could this ciphertext be mapped to plaintext? I understand a plaintext $m$ being mapped to numerous $c_i$ ?

Also to decrypt, don't we need to know $r$? In that case $r$ becomes a key
To decrypt the ciphertext you definitely need to recover the random token $r$.
But $r = Y \oplus H(X)$. Then knowing $r$ the padded message can be recovered as $m0\dots0 = X \oplus G(r)$. The decryptor only need to known the size of the random token $r$ and this is usually part of the standard. Of course the standard has to define the $F$ and $G$ functions and how to pad the message. The padding process has to be such that unpadding the message after the decryption is deterministic and unique. 

The randomness added to the encryption process makes impossible for an attacker having access to an encryption oracle (or observing a lot of messages and their effect) to create a dictionary of corresponding couples of $(m_i, c_i)_{i\in S}$ where $c_i = Enc(m_i)$. This property has been defined for the first time in the paper by Goldwasser and Micali "Probabilistic Encryption" (Shafi Goldwasser and Silvio Micali, Probabilistic Encryption, Special issue of Journal of Computer and Systems Sciences, Vol. 28, No. 2, pages 270-299, April 1984)