Intuition behind Gubinelli derivative

In a way it is very much like a usual derivative. Recall first that for a regular function $Y$, its derivative $Y'_s$ at a point $s$ is the (unique) number such that $$ Y_{t,s}=Y'_s(t-s)+ R_{s,t}, $$ where $R_{s,t}\to0$ faster than linearly. If $Y$ is twice differentiable, then $R_{s,t}\lesssim |t-s|^2$. That is, as a function of $t$, $Y_t$ "looks like" the linear function $Y_s+Y'_s(t-s)$, in the neighborhood of $s$.

Now simply replace the linear function by $X$. So we impose $$ Y_{t,s}=Y'_sX_{t,s}+R_{s,t} $$ with the remainder $R_{s,t}\to0$ faster than the first term, that is, faster than $|t-s|^\alpha$ (The condition $R_{s,t}\lesssim|t-s|^{2\alpha}$ from Friz-Hairer corresponds to the twice differentiable scenario in the previous case). Then as a function of $t$, $Y_t$ "looks like" the path $Y_s+Y'_sX_{s,t}$. This is great news for integration: we can of course integrate $Y_s$ against $dX_t$ (since as a function of $t$ it is just constant), and we can also integrate $X_{s,t}$ against $dX_t$ (by the definition of a rough path).

Actually, I wouldn't focus so much on assigning a meaning to $Y'$ itself, but rather focus on what the existence of a $Y'$ means for $Y$.

We want to define $\int_0^T f(X_s) dX_s$ for smooth bounded $f$ with bounded derivatives of all orders. Using linearity and a partition ${t_k}$ of $[0,T]$, we have

\begin{align*}\int_0^T f(X_s) dX_s&=\sum_k\int_{t_k}^{t_{k+1}}f(X_s) dX_s\\&=\sum_k\int_{t_k}^{t_{k+1}}f(X_{t_k})+f'(X_{t_k})(X_s-X_{t_k})+O(|s-t_k|^{2\alpha})dX_s\\&=\sum_k f(X_{t_k})(X_{t_{k+1}}-X_{t_k})+f'(X_{t_k})\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s+O(|t_{k+1}-t_k|^{3\alpha})\end{align*}

As $3\alpha>1$ the third term goes to zero as the mesh size goes to $0$. The first term is just a Riemann integral. The second term is the "rough path" term. $f'(X_{t_k})$ is the Gubinelli derivative and $\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s$ is your area process.