Let $E \xrightarrow{ \pi} M$ be a smooth vector bundle. If $E$ is the trivial bundle $ M \times \mathbb{R}^k$ then we can define directional derivatives of a section $\sigma \in \Gamma(E)$ (without needing a connection $\triangledown$) by $$D_X \sigma := \lim_{t \to 0} \frac{\sigma(\gamma(t))-\sigma(\gamma(0))}{t} \quad (*)$$ for some curve $\gamma$ in $M$ with $\gamma(0) = p$ and $\gamma'(0) = X_p$, since $\sigma$ can be identified with a map from $M$ to a fixed vector space $\mathbb{R}^k$.
My question is, since every vector bundle is locally trivial, then at least in some small neighbourhood of $p$ we can identify $\sigma(\gamma(t)) \in E_{\gamma(t)}$ and $\sigma(\gamma(0)) \in E_{\gamma(0)}$ with their images under the linear isomorphism $h_{\gamma(t)} : E_{\gamma(t)} \to \{\gamma(t)\} \times \mathbb{R}^k \cong \mathbb{R}^k$ where $h : \pi^{-1}(U) \to U \times \mathbb{R}^k$ and subtract them to define $(*)$.
Why is this approach not enough? I've read that the problem is that there is no natural isomorphism between $E_p$ and $E_q$ but I don't understand this statement very well. Any clarification would be great!
There is a little caveat here that you should look out for: The definition that you give is only right for the bunlde $M \times \mathbb{R}^k$, and not for any trivial bundle. That is because to make sense of your definition, you need a global trivialization first, which is canoncal in this case, but not in general. Your definition, i.e. the value of this directional derivative, depends on this in a crucial way, as you need to define some kind of horizontal direction on the tangent space of your bundle, which is exactly what a connection would do. But as long as you don't specify the trivialization you want to use, there usually is no canonical way of doing so, and this gets even worse for nontrivial bundles, where there is no global trivialization at all and the whole definition depends on the interchangability of local trivializations.
To conclude: You could of course use this definition, but if would (even for trivial bundles) depend on the non-unique choice of (local) trivialization, and it wouldn't be compatible at the intersection of two different local trivializations as well, so you do not even get a well-defined global section.