Proving the linearity of $df$: How to place the limit inside the argument of $\psi \circ f \circ \phi^{-1}$?

In the book of Chillingworth, the author defines the tangent space of a point $ p$ in the smooth manifold $ M$ as the set of all conjugacy classes of smooth paths with $ \alpha (o) = p$ s.t $ \alpha \sim \beta$ iff $ $ \lim_{t\to 0} \frac{\phi \circ \alpha(t) – \phi \circ \beta (t)}{t } = 0,$ $ where $ \phi$ is a local coordinate chart around $ p\in M$ .

Now, given a smooth map from $ f : M \to N$ , I’m trying to show that $ df: T_p M \to T_{f(p)} N$ given by $ [\alpha] \mapsto [f\circ \alpha]$ is a linear map.

However, to show that, I need to show that $ $ \lim_{t\to 0} \frac{\psi \circ f \circ \alpha(t) – \psi \circ f \circ \beta (t)}{t } = 0,$ $ where $ \psi$ is a local coordinate chart around $ f(p)$ .

In $ \mathbb{R}^n$ , I’m aware of this property, but even if I modify the limit as $ $ \lim_{t\to 0} \frac{\psi \circ f \circ \phi^{-1} \circ [\phi \circ \alpha(t) – \phi \circ \beta (t)]}{t } = 0,$ $ how to put the factor $ t$ in the denominator inside the argument of $ \psi \circ f \circ \phi^{-1}$ ?