If we have a curve $\gamma \circ f(t)$ (where $\gamma$ is a chart) then we can construct a tangent vector $$\xi = \frac{d}{dt}(\gamma \circ f(t))_{|t=0}$$ if we have another chart map, say $\tilde{\xi}$ then apparently we can relate it's induced tangent vector to the first one by $$\tilde{\xi}= D(\tilde{\pi}\circ \pi^{-1})_{|\pi p} \xi$$
I have no idea how to proceed except that it must require the chain rule. A good hint would be appreciated and then I'll try to post a solution.
A general way of achieving this is as follows, placed specifically in your example should give you a way to approach the setting.
One could define a space of vectors $T_x M$ in such a way that the map $D_x \varphi$ of the chart map $\varphi$ makes sense as a linear operator between vector spaces $T_x M$ and $\mathbb{R}^n$, and so that the chain rule continues to hold. Then we would have for any vector $v \in T_p M$ vectors $u = D_x \varphi (v) \in \mathbb{R}^n$, and $w = D_ x \eta(v) \in \mathbb{R}^n$.
Writing this another way (implicitly assuming the chain rule holds) we have $$D_\varphi(x)\varphi^{−1}(u) = v = D_\eta(x) \eta^{−1}(w).$$ The chain rule would then imply $$D_\varphi(x) (\eta \cdot \varphi^{-1})(u)=w$$