Probably a silly question, but the following seem to me intuitive definitions to make:
Throughout let $M$ and $N$ be smooth manifolds of dimension $m$ and $n$ respectively.
Definition: the tangent space of $M$ at $p\in M$ is $\mathbb{R}^m$, or $$\{p\}\sqcup \mathbb{R}^m$$ if you will (both spaces are isomorphic, so there is no relevant difference).
Intuition: given a chart $(U,\phi)$ at $p$, we see that $U$ is homeomorphic to an open subset $\phi(U)$ of $\mathbb{R}^m$, with $\phi(p)$ the image of $p$ under such homeomorphism. The natural tangent space at $\phi(p)$ is of course $\mathbb{R}^m$. Furthermore, if $T_pM$ is defined by means of derivations it can be shown that the derivations $$\left\{\frac{\partial}{\partial x_1}\Bigg|_p,\ldots,\frac{\partial}{\partial x_m}\Bigg|_p\right\}$$ form a basis of $T_pM$, which therefore is isomorphic to $\mathbb{R}^m$.
Definition: let $F:M\to N$ be a map between manifolds, with $p\in M$ and $F(p)=q\in N$. Given the charts $(U,\phi)$ and $(V,\psi)$ of $p$ and $q$ respectively, we say $F$ is differentiable at $p$ iff so is $$\psi\circ F\circ\phi^{-1}:\phi(U)\to\psi(V)$$ at $\phi(p)$; in such case we define the differential $dF_p$ of $F$ at $p$ as the differential of $\psi\circ F\circ\phi^{-1}$ at $\phi(p)$.
Intuition: again by 'interpreting' $p$ and $q$ as $\phi(p)$ and $\psi(q)$ respectively, and in their corresponding neighborhoods.
However, authors go through the trouble of defining the differential of $F$ in terms of derivations (or germs, or whatever objects were used to define the tangent spaces). Why is this the case?
The tangent space is isomorphic to $\mathbb R^n$ as you say, but not in a natural way. One way of thinking of this is in terms of continuity. If there were a continuous way of identifying each tangent plane of, say, the $2$-sphere $S^2$ with $\mathbb R^2$ (with basis $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$), this would imply the existence of an everywhere nonvanishing tangent vector field on $S^2$, where at every point we would choose the tangent vector corresponding to, say, $\frac{\partial}{\partial x}$. But we know that the $2$-sphere does not admit such a vector field by the hairy ball theorem.