Let $E\rightarrow M$ be a vector bundle with connection $\nabla$. Denote by $\Pi_{\gamma(t_{0})}^{\gamma(t_{1})}:E_{\gamma(t_{0})}\rightarrow E_{\gamma(t_{1})}$ the parallel transport map along the curve $\gamma$.
Show that for any $X\in \Gamma(TM)$ and $s\in\Gamma(E) $ $$(\nabla_{X}s)(p)=\lim_{\delta\rightarrow 0}\dfrac{\Pi_{\gamma(\delta)}^{\gamma(0)}(s(\gamma(\delta)))-s(\gamma(0))}{\delta}$$ where $\gamma$ is any curve in $M$ such that $\gamma(0)=p$ and $\gamma'(0)=X(p)$.
So I have no idea how to do this should any tips for this?
This shows you, that the connection is completely encoded in it's parallel transport and in this encoding, it's nothing than the regular derivative.
Let $\xi_1,...,\xi_n$ be a basis of $E_p$ and $s_1,...,s_n$ be the unique parallel sections along $\gamma$ with $s_i(0)=\xi_p$. Define $\tilde{S}=S\circ\gamma$ and let $\tilde{S}(t)=\sum\sigma^j(t)s_j(t)$. The following calculation then does the job:
\begin{align*} \nabla_{X_p}S = \tilde S'(0) &= \nabla_t\big(\sum_j\sigma^js_j\big)(0)\\ &= \sum_j\big((\sigma^j)'s_j + \underbrace{\sigma^j\nabla_ts_j}_{\equiv\ 0}\big)(0)\\ &= \sum_j\bigg(\lim_{t\to0}\frac{\sigma^j(t)-\sigma^j(0)}{t}\bigg)s_j(0)\\ &= \lim_{t\to 0}\frac{1}{t}\Big(\sum_j\sigma^j(t)s_j(0) - \sigma^j(0)s_j(0)\Big)\\ &= \lim_{t\to 0}\frac{1}{t}\Big(\sum_j\sigma^j(t)P_{t,0}^\gamma(s_j(t))-\sigma^j(0)s_j(0)\Big)\\ &= \frac{1}{t}\Big(P_{t,0}^\gamma\big(\sum_j\sigma^j(t)s_j(t)\big)-\sum_j\sigma^j(0)s_j(0)\Big)\\ &= \lim_{t\to0}\frac{P_{t,0}^\gamma(\tilde S(t))-\tilde S(0)}{t} \end{align*}
I think you can handle the few changes of notation. Otherwise feel free to ask.