Can we decrypt in this order when the message is encrypted twice? If we encrypt a message twice with symmetric key $k_1$ first and then $k_2$ like $E_{k2}\{E_{k1}\{m\}\}$ , ideally we should decrypt with $k_2$ first and then $k_1$ but is it possible to decrypt with $k_1$ and then with $k_2$ ?  Using any of the AES Modes preferably ?
What you are asking appears to be 'is AES commutative'?
The short answer to which is no: encrypting with AES with key 1 then key 2 will not (generally) give the same output as encrypting with key 2 then key 1, which is what would be required for naive implementation.

However, there are modes in which AES can be used which would be commutative.
For example, if you run AES in counter mode, then after encrypting with key1 and key2 through counter mode, we would have that the final cipher text $c_i$ was:
$$c_i = m_i \oplus E_{k_1}(i) \oplus E_{k_2}(i)$$

Since XOR is commutative, we can write this as either

$$c_i = [m_i \oplus E_{k_1}(i)] \oplus E_{k_2}(i) = [m_i \oplus E_{k_2}(i)] \oplus E_{k_1}(i)$$

That is, whichever order you encrypt (and thus whichever order you decrypt) with respect to the two keys, you will reach the same ciphertext (and thus the same plaintext).

I haven't looked at it in detail, but I think the following article might be of use to you:
"Is there two key symetric commutative encryption function?".

Edit: Just to point out, if one was to use counter mode in this way, you would have to either make sure the counters are initialised at different values, or that the keys are different - otherwise your 'encryption' would become
$$c_i = m_i \oplus E_{k_1}(i) \oplus E_{k_1}(i) = m_i$$
Which is probably not a good thing...