How is the tangent space given a vector space structure, when defined via smooth curves? As I understand it, the tangent space $T_{p}(M)$ to a manifold is given a vector space structure by taking a chart $\varphi:U\rightarrow V\subset\mathbb{R}^{n}$ and making the identification via the induced map $d\varphi_{p}:T_{p}(U)\rightarrow T_{\varphi(p)}(V)$, which is isomorphic to $\mathbb{R}^{n}$ (and this identification is independent of chart). How do we know that we can find a smooth map $\varphi$ for which $d\varphi_{p}$ is bijective?
 A: Recall that a chart about $p \in M$ is a pair $(U, \varphi)$, where $U \subset M$ is an open neighborhood of $p$ and $$\varphi: U \longrightarrow V \subset \mathbb{R}^n$$ is a diffeomorphism. Since $\varphi$ is a diffeomorphism, $$d\varphi_p : T_p U \longrightarrow T_{\varphi(p)} V$$ is a bijection (you can easily show that $d(\varphi^{-1})_{\varphi(p)}$ is its inverse). $\varphi$ exists because $M$ can be covered by charts.
A: There are some  subtle points here.
Given a chart $\phi :U\to V$, the bijection  $d\phi_p: T_p(M) \stackrel {\cong}{\to} \mathbb R^n= T_{\phi(p)}(V)$ is defined by sending the equivalence class of a curve $\gamma$ passing through $p$ at time $0$ to the velocity vector $(\phi\circ \gamma)'(0)= \vec u\in \mathbb R^n $.
 It is injective by the definition  of the equivalence relation on curves through $p$ and surjective by consideration of the curves  $t\mapsto \phi^{-1}(\phi (p)+t\vec u) \;( u\in \mathbb R^n)$
The difficulties are  that these bijections   $d\phi_p$ are very, very dependent on the chart $\phi$ and that they are not isomorphisms because $T_p(M)$ is  a priori a set and not a vector space ! So what is to be done ?
Here is what:   
Suppose you have two elements  $v,w\in T_p(M)$ . How do you add them?
The recipe is: choose a chart $\phi: U\to V$, compute $d\phi_p(v)=v'\in \mathbb R^n$ and $d\phi_p(w)=w'\in \mathbb R^n$, add these vectors in $\mathbb R^n$ and obtain $v'+w'$. Finally the required sum is defined as $$v+w\stackrel {def}{=}(d\phi_p)^{-1}(v'+w')$$
Yes, but what if you had chosen another chart $\psi: U\to W$?
You would have obtained $d\psi_p(v)=v''\in \mathbb R^n$, $d\psi_p(w)=w''\in \mathbb R^n$ and $v''+w''\in \mathbb R^n$ with:
 $v'$ very, very different from $v''$,
 $w'$ very, very different from $w''$
$v'+w'$ very, very different from $v''+w''$  
However you would observe with delight that $(d\phi_p)^{-1}(v'+w')=(d\psi_p)^{-1}(v''+w'') $ (very,very equal !)  so that the   definition of the sum of  $v, w \in T_p(M)$   $$v+w\stackrel {def}{=}(d\phi_p)^{-1}(v'+w')=(d\phi_p)^{-1}(d\phi_p(v)+d\phi_p(v))\in T_p(M)$$  does not depend on the choice of the chart $\phi$, despite strong appearances to the contrary !
An analogous observation will show you that the definition $r\cdot v\stackrel {def}{=}(d\phi_p)^{-1}(r\cdot d\phi_p(v))$ defines the product of a vector by a real scalar  and this  puts the final touch to the definition of the  vector space structure on $T_p(M)$ .
And so, finally, the answer to your question is:    
" For every chart $\phi$ the map  $d\phi_p: T_p(M)\to T_{p}(\mathbb R^n)$ is an isomorphism because it is a bijection and because we defined the vector structure on $T_p(M)$ in order that it be linear ! "
A: Here is a slightly different way to state what was already said in the other answers.
(Definition of tangent space via curves) Let the "tangent space $T_p M$ be defined as the set of equivalence classes of curves passing through $p\in M$ which, in any choice of coordinate chart, have the same slope. More precisely, let $\mathrm{SmoothCurves}(p)$ denote the set of $C^\infty(p)$ curves $\gamma:(-\epsilon,\epsilon)\to M$ for some $\epsilon>0$ such that $\gamma(0)=p$. We then say that $A\in T_p M$ if $A$ is a subset of curves, $A\subset\mathrm{SmoothCurves}(p)$, such that $\gamma,\eta\in A$ iff there is some local coordinate chart $(\phi,U)$ defined around $p$ such that $(\phi\circ\gamma)'(0)=(\phi\circ\eta)'(0)$.
We denote with $\gamma'(0)\equiv A$ the equivalence class containing the curve $\gamma$.
One can verify that such definition does not depend on the choice of chart $\phi$ (albeit the individual values of $(\phi\circ\gamma)'(0)$ do depend on $\phi$).
(Differential of a chart) For any $\gamma'(0)\in T_p M$ and chart $\phi:U\to \mathbb R^n$ with $p\in U\subseteq M$, we can now define the mapping $d\phi_p: T_p M\to T_{\phi(p)}\mathbb R^n\simeq \mathbb R^n$ as:
$$d\phi_p(\gamma'(0)) \equiv (\phi\circ\gamma)'(0).$$
Again, one can verify that the choice of $\gamma\in \gamma'(0)$ does not affect this definition.
This is indeed the differential of the chart, albeit at this stage one might not have properly defined the notion of differential on manifolds (depends on the treatise).
($d\phi_p$ is bijective) The differential map $d\phi_p$ is bijective.

*

*To see injectivity, suppose $d\phi_p(\gamma'(0))=d\phi_p(\eta'(0))$. This means that $(\phi\circ\gamma)'(0)=(\phi\circ\eta)'(0)$ for some (any) $\gamma\in \gamma'(0)$ and $\eta\in \eta'(0)$. But then, by the way the equivalence classes are defined, this implies that $\gamma'(0)=\eta'(0)$, as $\gamma$ and $\eta$ have the same slope with respect to some chart ($\phi$).

*To observe surjectivity, consider an arbitrary element $w\in T_{\phi(p)}\mathbb R^n$. In $\mathbb R^n$ we can always easily build a basis for the tangent space, writing $w=\sum_i c_i \gamma_i'(0)$ for a collection of curves
$$\gamma_i(t) \equiv \phi(p) + t \mathbf e_i,$$
where $\mathbf e_i$ is a basis for $\mathbb R^n$. Define then the curves $\tilde\gamma_i$ in $M$ as
$$\tilde\gamma_i(t) \equiv \phi^{-1}(\gamma_i(t)) = \phi^{-1}(\phi(p)+t \mathbf e_i)\in M.$$
Then
$$d\phi_p(\tilde\gamma_i'(0)) = (\phi\circ\tilde \gamma_i)'(0) = \gamma_i'(0).$$
Thus $w=d\phi_p(\sum_i c_i \tilde\gamma_i'(0))$, and $d\phi_p$ is surjective.

(Defining linear operations on $T_p M$) We therefore have a bijective map between $T_p M$ and $\mathbb R^n$. We can now leverage this to define the operations in $T_p M$ via their natural counterparts in $\mathbb R^n$: given any $\gamma'(0),\eta'(0)\in T_p M$ we define
$$\gamma'(0) + \eta'(0) \equiv (d\phi_p)^{-1}(d\phi_p(\gamma'(0)) + d\phi_p(\eta'(0))),$$
and similarly for scalar multiplication.
(Verifying robustness of the definition) One might notice that, in the above definition of $\gamma'(0)+\eta'(0)$, we made a specific choice of coordinate chart $\phi$. This is not ideal: we would like operations on $T_p M$ (which is not defined with respect to a specific choice of chart) to be coordinate-independent.
Fortunately, this turns out to be the case with this definition.
To see this, let $\psi:U_\psi\to\mathbb R^n$ with $p\in U_\psi\subseteq M$ be another chart. Define $\tau\equiv \psi\circ\phi^{-1}$.
Then we have
$$(d\psi_p)(\gamma'(0)) \equiv (\psi\circ\gamma)'(0)
= d\tau_{\phi(p)} \circ d\phi_p(\gamma'(0)).$$
Therefore,
$$d\psi_p(\gamma'(0)) + d\psi_p(\eta'(0))
= d\tau_{\phi(p)} \circ (d\phi_p(\gamma'(0)) + d\phi_p(\eta'(0))),$$
and finally
$$
(d\psi_p)^{-1}(d\psi_p(\gamma'(0)) + d\psi_p(\eta'(0)))
= (d\phi_p)^{-1}(d\phi_p(\gamma'(0)) + d\phi_p(\eta'(0))).
$$
