What is the relationship between entropy conditioning and final output bias in a TRNG?
The notation is a little confusing because it uses the same letter $n$ for two purposes, so let's change it to be a bit clearer:


  
  An entropy source output string containing $2t$ bits of entropy can be conditioned into a string of $t$ bits that contains full entropy output using an approved conditioning function, where $t$ is the length of the output block of the approved derivation function (see SP 800-90B).
  


(Replaced $n$ by $t$.)

In other words, if we have a physical device that produces a string $\sigma$ with $2t$ bits of entropy, which is necessarily at least $2t$ bits long, and we can pick a conditioning function $C\colon \{0,1\}^* \to \{0,1\}^t$, then the string $C(\sigma)$ is $t$ bits long and has nearly $t$ bits of entropy—in particular, $(1 - \varepsilon) t$, where $\varepsilon &lt; 2^{-64}$.

This is a statement of an assumption (in §4.2 ‘Assumptions’) about the entropy source, having at least $2t$ bits of entropy where $t$ is the length of the output block.


  
  The derivation functions in SP 800-90A distribute the entropy provided by the input string is $n$ bits (so that the length of the input string is $r$ bits for some $r \geq n$), and the length of the output string is $t$ bits, then the following is true:
  
  
  If $t \leq n/2$, then the output string has full entropy output (i.e., the output string has $(1 - \varepsilon) t$ bits of entropy, for some $\varepsilon \leq 2^{-64}$).
  
  


($t$ is the same as above now.)

In other words, if we have a string $\sigma$ with $n$ bits of entropy, which is necessarily at least $n$ bits long ($r \geq n$), then for any choice of derivation function $C\colon \{0,1\}^* \to \{0,1\}^t$ in SP 800-90A with $t \leq n/2$, the $t$-bit string $C(\sigma)$ has nearly $t$ bits of entropy—in particular, $(1 - \varepsilon) t$, where $\varepsilon &lt; 2^{-64}$.

This is a statement of an assumption about the derivation function, like Hash_df, used as a conditioning function.  An example of a function that would fail to satisfy this property is one that, say, just gives the first $t$ bits of the input string.

Why should these relations hold?  See an earlier answer for some details.  The quick summary is that a rough approximation for expected entropy of the output $F(\sigma)$ of a uniform random function $F\colon \{0,1\}^* \to \{0,1\}^t$ on a string $\sigma$ with $t + k$ bits of entropy is $\lg 2^t (1 - e^{-2^k}) = t + \lg (1 - e^{-2^k}) \approx t - e^{-2^k}/\log 2$, so as long as $k$ is reasonably large we easily attain ‘full entropy’ in this model with $\varepsilon \approx e^{-2^k}/(t\log 2)$.

Picking $k \geq 64$ is plenty—and we are required to do that because the block size of every approved derivation function is at least 64.  We don't have a uniform random function in the real world, but the derivation functions of SP 800-90A don't seem to have any interesting properties that let us distinguish them from one, so it's a pretty good model.

What if you chose a string $\sigma$ with $n = 256$ bits of entropy, and a 3DES-based derivation function with an output block of $t = 64 = n/4$ bits?  Then $k = 192$, so $\varepsilon$ is really quite tiny, and we get ‘full entropy’ in a 64-bit string—although that's only just under 64 bits of entropy.

What if you chose a string $\sigma$ with $n = 130$ bits of entropy, and an AES-based derivation function with an output block of $t = 128$ bits?  The margin is smaller in this case: since $k = 2$, the approximation above (a) is not a very good approximation and (b) gives an $\varepsilon$ nowhere near $2^{-64}$.  Maybe we get more than 64 bits of entropy in the end, but that's not ‘full entropy’ for a 128-bit string.

At what $k$ do we draw the line?  To give a comfortable margin without having to think hard about it, we just say that you should use twice as much the entropy in the input as the ${\geq}64$-bit output is long, in order to reliably get ‘full entropy’ in the output.