Composition of block ciphers and 3DES
Well, whether $AES'$ is as secure as $AES$ depends on the length of $k_1, k_2$.

If they are both 128 bit, then what you effectively have is a standard 128-bit AES, except that prior to round 6, you replace the running key with an independent key (and you tweaked the last round, but that's cryptographically harmless).  Now, it is never a good idea to do random tweaks on ciphers, however, this appears reasonable to conjecture that this one is harmless; that the security of the cipher does appear unlikely to depend on the relationship of the subkeys in round 5 and round 6 of standard AES.  This, of course, assumes that we are interested solely in the standard attacks (chosen-plaintext, chosen-ciphertext); if we are allowed to consider related-key attacks, this construction may well be weaker.

On the other hand, if $k_1, k_2$ are each 64 bits (with the $AES^r$ subkey derivation process being somehow modified to take a 64 bit key), then what we have is definitely weaker than standard AES (even though $AES'$ takes a 128 bit key); even against a known-plaintext scenario.

This is due to a standard meet-in-the-middle attack, where to find the values $k_1, k_2$ that solve the equation:

$AES'_{k1,k2}(P) = AES^r_{k1}(AES^r_{k2}(P)) = C$

for some known plaintext/ciphertext blocks, the attacker just computes:

$F(k_1) = AES^r_{k2}(P)$

$G(k_2) = {AES^r_{k1}}^{-1}(C)$

for all $2^{64}$ values of $k_1, k_2$, and searches the lists to find a pair $F(k_1) = G(k_2)$.  This gives him solutions for $AES'_{k1,k2}(P) = C$ in $O(2^{64})$ time.