Let $f: \mathcal{M} \to \mathcal{N}$ be a $\mathscr{C}^{r+1}$ map. We define a map $\mathscr{T}f: \mathscr{T}\mathcal{M} \to \mathscr{T}\mathcal{N}$ as follows:
A local representation of the map $\mathscr{T}f$ in the charts on $\mathscr{T}\mathcal{M}$ and $\mathscr{T}\mathcal{N}$ is simply the derivative of the local representation of $f$ in the charts $\{\psi_i,U_i\}_{i \in \Lambda_1}$ and $\{\theta_i, V_i\}_{i \in \Lambda_2}$ on $\mathcal{M}$ and $\mathcal{N}$, respectively. We have the requirement that $f(U_i) \subset V_i$. Thus we have a $\mathscr{C}^r$ map
$$ (\mathscr{T}f)_{ij}: \mathscr{T}U_i \to \mathscr{T}V_i$$
$$ [x,i,a] \mapsto [f(x),j,\partial(\theta_jf\psi^{-1}_i)(\psi_i(x))a] $$
Since this is independent of $i,j$, there is a well-defined map $\mathscr{T}f: \mathscr{T}\mathcal{M} \to \mathscr{T}\mathcal{N}$, so that if $f(x) = y$, $\mathscr{T}_xf: \mathscr{T}_x\mathcal{M} \to \mathscr{T}_y\mathcal{N}$.
Now my two questions are, why must we require that $f(U_i) \subset V_j$ for our open coverings? Also, why is the map $(\mathscr{T}f)_{ij}$ independent of $i,j$? I'm told that this is an application of the chain rule, but I don't quite see it.
Answer to the first one is technical. If we would take $U_i$ too big, then the map defined as it has been done $$(\mathscr{T}f)_{ij}: \mathscr{T}U_i \to \mathscr{T}V_j$$ $$[x,i,a] \mapsto [f(x),j,\partial(\theta_jf\psi^{-1}_i)(\psi_i(x))a],$$
wouldn't have much sense, because for $x\in U_i\setminus f^{-1}(V_j)$ point $y=f(x)$ wouldn't lie in $V_j.$ However if our covering $\{U_i\}$ would be arbitrary, then by the fact that $f$ is continuous we could find a new one which would suffice. Just take covering $\{U_i\cap f^{-1}(V_j)\}$ running for all $i,j.$
To the second one fix $x\in U_{i_1}\cap U_{i_2}$ and recall that the $[x,i_1,a_1]=[x,i_2,a_2]$ if $\partial(\psi_{i_2}\circ\psi_{i_1}^{-1})(\psi_{i_1}(x))a_1=a_2.$ Hence on the one hand $$[x,i_1,a_1] \mapsto [f(x),j,\partial(\theta_jf\psi^{-1}_{i_1})(\psi_{i_1}(x))a_1]$$ and on the other hand $$[x,i_2,a_2] \mapsto [f(x),j,\partial(\theta_jf\psi^{-1}_{i_2})(\psi_{i_2}(x))a_2].$$ By the mentioned fact we have that $\partial(\psi_{i_2}\circ\psi_{i_1}^{-1})(\psi_{i_1}(x))a_1=a_2.$ Plug it in the second, use chain rule and you have the first one. Similar trick with $j.$