I am trying to understand the following derivation in my lecture notes. Given an n-dimensional differentiable manifold $M$ and a parametrized curve $\gamma : (-\epsilon, \epsilon) \rightarrow M : t \mapsto \gamma(t)$, with $\gamma(0) = \mathbf{P} \in M$.
Also define an arbitrary (dummy) scalar field $f = C^{\infty}(M)$, and $$\bar{\gamma} = \phi_\alpha \circ \gamma, \qquad \qquad \bar{f} = f \ \circ \phi_\alpha^{-1}.$$ Here $\phi_\alpha(\mathbf{P}')$ gives the coordinates of the point $\mathbf{P'}$ on the manifold, and $\phi_\alpha^{-1}$ is its inverse. The tangent vector of $\gamma$ at $\mathbf{P}$ is then given as follows, $$\dot{\gamma}(0)f = \frac{d}{dt}(f \ \circ \gamma)|_{t=0} =\frac{d}{dt}(\bar{f} \circ \phi_\alpha \circ \phi_\alpha^{-1} \circ \bar{\gamma})|_{t=0} = \frac{d}{dt}(\bar{f} \circ \bar{\gamma})|_{t=0} = \dot{x}^i(0) \frac{\partial\bar{f}}{\partial x^i}|_{x(0)}.$$
Here $\dot{x}^i(t) = \bar{\dot{\gamma}}^i(t)$ and $x^i(t) = \bar{\gamma}^i(t)$ are coordinates. I understand everything up to this point. Next however the following step is introduced, $$\dot{x}^i(0) \frac{\partial\bar{f}}{\partial x^i}|_{x(0)} = \dot{x}^i(0) \frac{\partial f}{\partial x^i}|_{\gamma(0)}.$$
I cannot understand or derive why this holds. The lhs makes sense to me, since $\bar{f}$ takes coordinates as its arguments, and thus differentiating to coordinates seems logical. However on the rhs we are suddenly differentiating $f$, which takes a point on the manifold $\mathbf{P}$ as its argument. How can we still derive this to the coordinates? Trying to fill in the definitions of $\bar{f}$ and $\bar{\gamma}$ on the lhs also doesn't seem to take me anywhere. Any help would be greatly appreciated.
EDIT: After searching around a little more it seems that the equality that confuses me is actually a definition, not something that is derived. This however doesn't really lift my confusion. If it is just a definition, what are you actually defining? Everything holds for arbitrary functions $f$ and arbitrary points $\mathbf{P}$. So why can you simply say that this holds without further argument?