To answer the question you asked (more or less, "Can I swap the limit and the integral?"), I have no idea. There are some cases where you can, and I can never remember the rules. I had a student who called such swaps "engineer's prerogative," but I think this is a little unkind to engineers. I'm sure others can point you to theorems. But often it's easier just to do things by hand.
If you instead look at
\begin{align}
C(x) - C(x_0)
&= \frac{1}{2\pi} \int_0^{2\pi} f(x-t)g(t)~dt - \frac{1}{2\pi} \int_0^{2\pi} f(x_0-t)g(t)~dt\\
&= \frac{1}{2\pi} \int_0^{2\pi} [f(x-t) - f(x_0 - t)]\cdot g(t)~dt\\
&= \frac{1}{2\pi} \int_0^{2\pi} [f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)~ds
\end{align}
you can estimate the integrand: Since $f$ is continuous on the compact set $[0,2\pi]$, it's uniformly continuous. So for $\epsilon > 0$, there's a $\delta$ such that $|p - q| < \delta$ implies $|f(p) - f(q)| < \epsilon$. Furthermore, because $g$ is integrable, so is $|g|$ (explanation.) That means that for $|x - x_0| < \delta$, we have
\begin{align}
| C(x) - C(x_0) |
&= \frac{1}{2\pi} |\int_0^{2\pi} [f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)ds| \\
&\le \frac{1}{2\pi} \int_0^{2\pi} |[f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)|ds \\
&= \frac{1}{2\pi} \int_0^{2\pi} \epsilon \cdot |g(x_0 + s)|ds \\
&= \epsilon \frac{1}{2\pi} \int_0^{2\pi} |g(x_0 + s)| ds \\
&= \epsilon \frac{1}{2\pi} \int_0^{2\pi} |g(s)| ds \text{, by periodicity}\\
&= \epsilon M
\end{align}
where $M$ is the integral of $|g|$.
So as $\epsilon$ goes to zero, the difference goes to $0$ and $C$ is continuous at $x_0$.
