I am struggling to understand the definition of a tangent space $T_p M$ at a point $p$ in an $n$-dimensional smooth manifold $M.$ I will write what I think is a proper definition below, and it would be great if somebody could tell me if this definition is correct, or where I am going wrong. My understanding is very basic, so I am hoping the responses will be forgiving and simple:
Let $(U,\phi)$ be a chart with $p \in U.$
We call a curve $(-1,1) \overset{\gamma}{\rightarrow} M$ a differentiable curve initialized at $p$ when $\gamma (0) = p$ and $\phi \circ \gamma$ is differentiable. Let $\gamma_1$ and $\gamma_2$ be two differentiable curve initialized at $p.$ We say $\gamma_1$ and $\gamma_2$ are equivalent at $0$ (written $\gamma _1 \sim \gamma _2$) if and only if the $\phi \circ \gamma_1$ and $\phi \circ \gamma_2$ have the same derivatives at $0.$ This defines an equivalence relation, and we can consider the tangent space $T_p M$ to consist of the equivalence classes of differentiable curve initialized at $p$ under the equivalence relation $\sim$.
We write $\gamma ' (0)$ to denote the equivalence class of a differentiable curve $\gamma$ initialized at $p.$ Using a chart $(U,\phi)$ we define $T_p M \overset{d \phi _p}{\rightarrow} \mathbb{R}^n$ so as to send $\gamma ' (0)$ to the derivative of $\phi \circ \gamma,$ evaluated at $0$ (where $\gamma$ is in $\gamma ' (0)$). Now (and this is the part I am most unsure about) $d \phi _p$ is a bijection, and so we can think of $T_p M$ is a vector space because we can apply $d \phi _p$ to transform members of $T_p M$ to $\mathbb{R}^n,$ and then define vector operations there, and then use the inverse of $d \phi _p$ to transform back again. In particular, for $u, v \in T_p M$ we can define the sum of $u$ and $v$ to be $\left( d \phi _p \right)^{-1} \left( d \phi _p (u) + d \phi _p (v) \right), $ and we can define the multiplication of $u \in T_p M$ by a scaler $r \in \mathbb{R}$ to be $\left( d \phi _p \right)^{-1} \left( r. d \phi _p (u) \right).$