Well, the first thing comes to mind is "what if your 'read-only location' isn't quite as read-only as you had hoped; if someone could modify your $f(i) \oplus k_i$ share, could they modify the reconstructed shared secret in a controlled way.
In your first example, I believe they could. Let us assume that we are doing Shamir's Secret Sharing over the field $GF(2^k)$; let us further assume that the attacker has $N-1$ shares (where $N$ is the number of shares we need to reconstruct the secret). Which his $N-1$ shares, he can reconstruct that, when you add in your share $f(i)$, the reconstructed secret will be $s = c \times f(i) + d$, for some values $c, d$ he can compute (and $\times$ and $+$ are over $GF(2^k)$.
Now, $c \neq 0$, and so this doesn't give him any information about what the secret is (given that he doesn't know your share). However, if he wants to modify the secret, say, add $t$ to is, what he can do is modify your encoded share on the read-only location, replacing $(i, f(i) \oplus k_i)$ with $(i, f(i) \oplus k_i \oplus (t \times c^{-1}))$.
Then, when you reconstruct the secret, using his N-1 shares, and your modified share, you will decode your share as $F(i) + (t \times c^{-1})$ (remember, in $GF(2^k)$, addition and exclusive-or are the same operation); then, when you combine this decrypted share with his, you're come up with $c(F(i) + tc^{-1})+d = cF(i)+d + t = s+t$, resulting in the modified shared secret that the attacker selected.
On the other hand, by using AES to encrypt the share, the attacker cannot do any such tricks. He can modify the share so that the shared secret becomes something else, however, he has no control over what that something else might be.
