From the notes I am using,
For each point $p \in M$, the tangent space to $M$ at point $p$ is the set
Where,
Let $(M,O,A)$ be a smooth manifold. Let the curve $\gamma: \mathbb{R} \to M$, which is atleast $C^1$. Suppose $\gamma(\lambda_o)=p$. The velocity of $\gamma$ at the point $p$ of the curve is the linear map:
$v_{\gamma,p}:C^{\infty} (M) \to \mathbb{R}$
with, $v_{\gamma,p}(f)= (f \circ \gamma)'(p)$
where,$C^{\infty}(M) := \{ f: M \to R| \text{f is a smooth function }\}$
equipped with $(f + g)(p) = f(p) + g(p)$ and $(\lambda g)(p)= \lambda >\cdot g(p)$
$$T_p M := \{v_{\gamma,p }| \text{ for all smooth curves } \gamma \text{ through p }\}$$
With an addition and s multiplication defined the following way,
$$+:T_p M \times T_p M \to Hom(C^{\infty} (M), \mathbb{R})$$
$$(v_{\gamma,p} + v_{\delta,p} )(f) =v_{\gamma,p}(f)+ v_{\delta,p}(f)$$
$$ \cdot: \mathbb{R} \times T_p M \to Hom(C^{\infty} (M),R)$$
$$(\alpha \cdot v_{\gamma,p})(f):= \alpha \cdot v_{\gamma,p} (f)$$
In the standard vector space deifinition the addition maps into another element in the vector space itself. So, if $T_pM$ was a vector space then the addition must again map to $T_pM$ but here it doesn't, yet we call it a vector space, why?
Best Answer
First of all your definition of $T_p M$ is incomplete, it should be $$ T_p M = \{ \text{all smooth curves $\gamma$ through $p$} \}/\sim $$ where two curves $\gamma$ and $\eta$ are equivalent under $\sim$ if their derivatives at $p$ (after you compose with a chart) are equal.
There is a natural injection $$ i:T_p M \hookrightarrow \text{hom}(C^\infty(M), \mathbb{R})$$ where for a smooth curve $\gamma$ with $\gamma(0) = p$ and $f \in C^\infty(M)$ we have $$ i([\gamma]) (f) := (f \circ \gamma)'(0)$$.