Let $v \in T_pM$ for some smooth manifold,assume $v_n \in T_pM$ also.
Assume further that $\lim_n v_n = v$.
Do we have the identity that $$v_n(f) \to v(f)$$ for $f\in C^\infty(M)$
(I was tring to provd the relation which comes from Lee's smooth manifold book page 231:
$$\left.\frac{d}{d s}\right|_{s=0} d\left(\theta_{-t_{0}}\right) \circ d\left(\theta_{-s}\right)\left(W_{\theta_{s}}\left(\theta_{t_{0}}(p)\right)\right. =d\left(\theta_{-t_{0}}\right)\left(\left.\frac{d}{d s}\right|_{s=0} d\left(\theta_{-s}\right)\left(W_{\theta_{s}}\left(\theta_{t_{0}}(p)\right)\right)\right)$$
Where $W$ is a vector field over M.
that is same to prove(notation may differ but the idea is the same): $$\frac{d}{ds} d\theta_{t_0} (v(s)) = d\theta_{t_0}(\frac{d}{ds} v(s)) \tag{*}$$
I try to do it as follows (using definiton of differential) denote $\Delta v(s) = \frac{v(s) - v(0)}{s}$ write everything in coordinate:
$$\lim_s d\theta_{t_0} (\Delta v(s)) = \lim_{s\to 0}\sum d\theta_{t_0} (\Delta v(s))(x^i) \partial_i = \sum \lim_{s\to 0}d\theta_{t_0} (\Delta v(s))(x^i) \partial_i $$
Then $$\sum \lim_{s\to 0}d\theta_{t_0} (\Delta v(s))(x^i) \partial_i = \sum \lim_{s\to 0}[(\Delta v(s))(x^i\circ\theta_{t_0})] \partial_i$$
If we can show $$\lim [\Delta v(s) (f)] = (\lim \Delta v(s)) f$$ then we are done
since $$ \sum \lim_{s\to 0}[(\Delta v(s))(x^i\circ\theta_{t_0})] \partial_i = \sum d\theta_{t_0}[(\lim_{s\to 0})(\Delta v(s))](x^i) \partial_i =\sum d\theta_{t_0}(\frac{d}{ds} v(s))(x^i) \partial_i = RHS$$
From the fact that every smooth manifold can be endowed with a Riemannian metric (think of this as a scalar product on each tangent space that changes smoothly along the points of the manifolds), then we can define from it a Riemannian norm and hence a Riemannian distance. As a result, a notion of limit in each tangent space pops up. With this information, all should be reduced to the familiar case but some technical changes.