Why is a vector space equal to its tangent space for any point?

Why is a vector space equal to its tangent space for any point?

I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of
section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$,
then $T_x(V)=V$ if $x\in V$.
If $x\in V$, then since $V$ is a manifold, there is a local
parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$.
Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is
defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.
An arbitrary element of $T_x(V)$ looks like $$ d\phi_0(v)=\lim_{t\to
0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t} $$
but this doesn't seem very useful to show $T_x(V)=V$. What is the right
approach?