We have 3 different (but equivalent) definition of tangent space, one of them is by equivalent class of smooth path, i.e. let $c:(-1,1)\rightarrow M$ be a path on $M$ with $c(0)=p$, then $c_1\sim c_2$ if for every $f:M\rightarrow\mathbb{R}$, $(f\circ c_1)'(0)=(f\circ c_1)'(0)$.
However, when applying it to find the tangent space of north pole of $S^2$, I meet some trouble.
Consider path on $S^2$, $r(t)=(\text{sin}(\theta)\text{cos}(\varphi), \text{sin}(\theta)\text{sin}(\varphi), \text{cos}(\theta))$ where $\theta$ and $\varphi$ are smooth function of $t$.
Then for any $f:S^2\rightarrow \mathbb{R}$, we can regard $f$ as the function of the coordinates of the point on $S^2$,i.e. $f(P)=f(x,y,z)$, then we have $(f\circ r)'=f_xx'+f_yy'+f_zz'$ where $(x(t),y(t),z(t))=r(t)$.
So by definition, for two path to be equivalent, we must have their $x'$, $y'$, $z'$ are equal.
After calculation and set $\theta(0)=0$ (we can't set $\varphi(0)$ because it can be anything.), we get for $r_1$ and $r_2$ to be equivalent, we must have
$ \text{cos}(\varphi_1(0))\theta_1'(0)=\text{cos}(\varphi_2(0))\theta_2'(0) $ $ \text{sin}(\varphi_1(0))\theta_1'(0)=\text{sin}(\varphi_2(0))\theta_2'(0) $
From which we can get 2 solutions:
(1)$\theta_1'(0)=\theta_2'(0)\neq 0$$\varphi_1(0)=\varphi_2(0)$
(2)$\theta_1'(0)=\theta_2'(0)= 0$
I think the solution is quite strange, because it does not depend on $\varphi'(0)$ and the second condition doesn't depend on $\varphi(0)$. And if $\theta(t)=t^2$, then whatever $\varphi(t)$ is, those path are equivalent. It is obviously anti-intuition.
So, doesn't the tangent space at north pole be $\mathbb{R}^2$ like the intuition suggests?