In OFB, with $O_0=\mathrm{IV}$ and plaintext/ciphertext blocks $P_i/C_i$ numbered starting from $1$, we have block cipher encryption output $O_i=E_K(O_{i-1})$ and $C_i=E_i\oplus O_i$.
I am having trouble with is understanding how same plaintext at same position can be used to extract encryption output.
In cryptanalysis, the default is to assume that all ciphertext is known, and it's common to assume that the full plaintext of past messages is known. I assume that until the end of this first section.
Additionally knowing that a new plaintext block $P_i$ is identical to a past plaintext block thus reveals that new plaintext block, and then a XOR of $P_i$ and it's corresponding ciphertext block $C_i$ in the new message gives the encryption output $O_i$ in the new message.
We can similarly determine all $O_j$ for the whole of past messages. If any of these matches $O_i$ and the key $K$ did not change, then we can find $j$ and for which past message, and thus most often $O_{i'}$ for some $i'$ around $i$, since $O_{i'}$ in the new message is $O_{j+i'-i}$ in the old message with the colliding $O_j$. That lets us decipher some of the new message around block $i$ (from block $\max(i-j+1,1)$ to block $\min(\ell-i,m-j)+i$ where the new message has $\ell$ blocks and the message with $O_j$ has $m$ blocks, or something on that tune).
Note: it would be wrong to assume that the possibility of collision among the $O_i$ of various messages becomes sizable past $2^{b/2}$ blocks encrypted with the same key for a $b$-bit cipher (because cycles in iterated random bijections are typically much longer than for iterated random functions). The number of messages encrypted must also be considered. Even though, it's still dangerous to some degree to use the same key to encipher anywhere close to $2^{b/2}$ blocks (or $b\,2^{(b-25)/2}$ bytes if we want probability of collision less than one in a million), even if $\mathrm{IV}$ changes. That's a practical consideration for DES ($b=64$), much less so for AES ($b=128$).
one of the vulnerability mentioned for [OFB] is that if two different messages have a block at the same position in the ciphertext, and have same plaintext, the attacker can figure out the encryption function output for that particular block. This was brought up to highlight the danger of reusing IV, so this is assuming that same IV and key are used.
As pointed above, there is some (moderate) degree of danger even if $\mathrm{IV}$ did change. We are able to confirm that with low probability of false positive if we find two ciphertext blocks identical at the same index, which most likely is because plaintext blocks are identical at the same index, which does happen in practice.
Now assuming $\mathrm{IV}$ did not change across a few messages (regardless of how and even if that is confirmed), and all ciphertext but no known plaintext is available, an attacker stands a fair chance to decipher at least in part a lot of the plaintext, if it's redundant enough (e.g. known to be English in ASCII), much like the One-Time-Pad becomes insecure if the pad gets reused and the plaintext is redundant enough.
Update per comment: The above was written before the source of the problem was known. That problem makes reference to this text in the source, about OFB:
…the sequence of encryption output blocks, $O_i$, depends only on the key and the IV and does not depend on the plaintext. Therefore, for a given key and IV, the stream of output bits used to XOR with the stream of plaintext bits is fixed. If two different messages had an identical block of plaintext in the identical position, then an attacker would be able to determine that portion of the $O_i$ stream.
If we do not assume that the "identical block of plaintext" is somewhat known, that statement is wrong, as well as exercise 6.9 which asks how that determination can be made (it can not).
The only way to make sense of the statement and exercise is if we assume one of the messages is known (which is a common hypothesis, as stated in the first part of my answer). Then the exercise is trivial, but the statement quoted above is at best strange, since it assumes a fixed IV, and we don't need that hypothesis. The only way a fixed IV helps is that it allows recognition (with low probability of false positive) that two different messages have an identical block of plaintext in the identical position.
