Why do these two calculations give me different answers for the same acid-base titration?
Part of the reason why there is such a large discrepancy is because in your second method of calculating $[\ce{HA}]$, you failed to take into account the fact that your vinegar was diluted from $3.1\mathrm{~cm^3}$ to $50\mathrm{~cm^3}$. 

Effectively, if the original concentration of $\ce{HA}$ was $x$, then your dilution would have brought it to a concentration of $\frac{3.1}{50}x$. And you plugged the $\mathrm{pH}$ of this diluted solution into the the $K_\mathrm{a}$ equation. Let's adjust for that dilution:

$$\begin{align}
\frac{3.1}{50}x &amp;= 0.041\mathrm{~mol~dm^{-3}} \\
x &amp;= 0.661\mathrm{~mol~dm^{-3}}
\end{align}$$

As you can see, there is still some difference between the two values. However, you can calculate the theoretical $\mathrm{pH}$ of a $1.067\mathrm{~mol~dm^{-3}}$ solution of $\ce{CH3COOH}$, using the fact that the $\mathrm{p}K_\mathrm{a}$ is $4.6$. Here, I will make the simplifying assumptions that the final and initial concentrations of $\ce{HA}$ are the same, i.e. $[\ce{HA}] = c_{\ce{HA}}$, and also that $[\ce{H+}] = [\ce{A-}]$:

$$\begin{align}
\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} &amp;= 10^{-4.6} \\
&amp;= 2.511\times 10^{-5}\mathrm{~mol~dm^{-3}} \\
[\ce{H+}] &amp;= \sqrt{(2.511\times 10^{-5})(1.067)}\mathrm{~mol~dm^{-3}} \\
&amp;= 0.00518\mathrm{~mol~dm^{-3}} \\
\mathrm{pH} &amp;= 2.28
\end{align}$$

You could go all the way and do mass-balance and charge-balance equations in order to avoid making that assumption, but I doubt it alters the result significantly. So, if you really had a concentration of $1.067\mathrm{~mol~dm^{-3}}$, your observed initial $\mathrm{pH}$ would not have been $3.0$. There is likely some experimental error present, or perhaps the $\mathrm{pH}$ meter that you used to measure the initial $\mathrm{pH}$ is not calibrated well.