You don't need the continuity of $f$.
Proposition: If $f \in L^1(\Bbb{T})$ and $f(0^-)$ and $f(0^+)$ both exist then
$$
\sigma_nf(0) \to \frac{f(0^-) + f(0^+)}{2}\quad(n\to\infty)
$$
Proof: Let
$$
g(t) := \begin{cases} \phantom{-}1-t/\pi & \phantom{-}0 < t \leq \pi \\ -1+t/\pi & -\pi \leq t < 0 \\ \phantom{-}\phantom{-}0 & \phantom{-}t = 0\end{cases}
$$
and
$$
\tilde{f}(t) := \begin{cases} f(t)-\frac{f(0^+)-f(0^-)}{2}g(t) & t \neq 0 \\ \phantom{-}\phantom{-}\phantom{-}\frac{f(0^+)-f(0^-)}{2} & t = 0 \end{cases}
$$
Then $\tilde{f} \in L^1(\Bbb{T})$ and $\tilde{f}$ is continuous at $0$.
Note that $g$ is an odd function hence $\hat{g}(-k)=-\hat{g}(k)$ for all $k \in \Bbb{Z}$. In particular,
$$
s_ng(0) = \sum_{k=-n}^n \hat{g}(k)=0
$$
so
$$
\sigma_ng(0)=0\quad(n\geq0)
$$
Hence
\begin{align}
\sigma_nf(0) &= \sigma_n\tilde{f}(0) \\
&\to \tilde{f}(0) & \text{by Fejér's theorem}\\
&=\frac{f(0^-) + f(0^+)}{2}
\end{align}
as $n\to\infty$.