If $g$ is $C^1$ on the unit circle $C(0,1)$. Then there is a function $f^+$ holomorphic on $B(0,1)$ and continuous on $\bar B(0,1)$, a function $f^-$ holomorphic on $\mathbb{C}\backslash\bar B(0,1)$ and continuous on $\mathbb{C}\backslash B(0,1)$, such that $g(z)=f^+(z)-f^-(z)$ on the unit circle.
This seems like it could be a direct consequence of some of the lecture results (e.g. Cauchy's Integral Formula, Taylor series), but cauchy's formula just gives a holomorphic function on $B(0,1)$, I do not know how that relates to the value of $g(z)$. While Taylor series only applies for functions holomorphic on some domain. If we can extend $g$ to some holomorphic function on $\mathbb{C}$ then surely the result follows... But how do we do something like that? Any help will be appreciated.
It is possible. Write the Fourier series of $g$ on the unit circle as
$$ g(z)=\sum_{n\in\mathbb Z} a_n z^n. $$
This gives the usual Fourier series for those $z$ such that $|z|=1$, and the series on the right does not necessarily converge for the other values of $z\in\mathbb C$. Since $g$ is $C^1$, the Fourier coefficients $(a_n)_{n\in\mathbb Z}$ decay rapidly (we have at the very least $(na_n)_{n\in\mathbb Z}\in \ell^2(\mathbb Z)$). Now let
$$ f^+(z):=\sum_{n\geq 0} a_n z^n, $$
$$ f^-(z):=\sum_{n<0} a_n z^n. $$
Both series converge uniformly to continuous functions $f^+$ and $f^-$ on respectively $\bar B(0,1)$ and $\mathbb C\setminus B(0,1)$ due to the fast decay of the coefficients $a_n$. It is not hard to show holomorphicity of the two functions in the desired sets.
As you can see, the only choice you can make is whether to include the term $a_0$ in $f^+$ or in $f^-$. The rest of the terms in the Fourier series are slpit into the two functions depending on where they are singular. The maps $g\mapsto f^+$ and $g\mapsto f^-$ are sometimes called projection maps onto positive and negative frequencies respectively. One has similar maps for other spaces, most notably $L^2(\mathbb T)$, $L^2(\mathbb R)$, the corresponding Sobolev spaces, etc...