Perfect Secrecy for two distinct messages
Let $m_i := 0$ and $k_i := \langle 0, 1 \rangle$. Then $c_i = Enc_{k_i}(m_i) = m_i \oplus k_{i0} = 0\ \oplus\ 0 = 0$. But then $m'_i = Dec_{k_i}(c_i) = c_i \oplus k_{i1} = 0\ \oplus\ 1 = 1$, and we have $m_i \neq m'_i$. The same holds for $m_i := 1$ and $k_i := \langle 1, 0 \rangle$. So your encryption scheme might give perfect secrecy, but it does not allow you to decrypt the message. 

This is because your encryption scheme effectively loses information from the message when encrypting with a key of $k_i = \langle 0, 1 \rangle$ or $k_i = \langle 1, 0 \rangle$, and for these keys the ciphertext $c_i$ is solely determined by the key:

$Enc_{\langle 0, 1 \rangle}(0) = 0\ \oplus\ 0 = 0\\
Enc_{\langle 0, 1 \rangle}(1) = 1\ \oplus\ 1 = 0\\
Enc_{\langle 1, 0 \rangle}(0) = 0\ \oplus\ 1 = 1\\
Enc_{\langle 1, 0 \rangle}(1) = 1\ \oplus\ 0 = 1$

If you reduce the keyspace to $\mathcal{K} = \{\langle 0, 0 \rangle, \langle 1, 1 \rangle\}^l$ you effectively have a regular one-time pad.