If I have an open subset $U$ of a n-dimensional $C^k-$manifold $M$ and a homeomorphism $f:U \to \Bbb R^n$ (Basically I am talking about a chart $(U,f)$) can I say that under this map a basis of $T_pU=T_pM$ can be sent to a basis of $T_{f(p)}\Bbb R^n$ and even vice versa? that is if we have a basis of $T_{f(p)}\Bbb R^n$ can it be pulled back to a basis of $T_pM$? Here $p$ is a point inside $U$ and I am actually talking about the differential $df_p:T_pM \to T_{f(p)}\Bbb R^n.$ So I am seeking answer in the context of differentials only.
By a n -dimensional $C^k-$manifold, I mean a topological manifold with a maximal $C^k-$atlas i.e we have a collection of charts $\{(U_i,f_i)\}$ forming a $C^k-$atlas where each $f_i : U_i \to \Bbb R^n$ is a homeomorphism and the transition maps are $C^k-$maps i.e. they are k-times continuously differentiable. By compatibility condition on $C^k-$atlases (Two $C^k-$atlases are compatible if their union is a $C^k-$atlas), we have a maximal $C^k-$atlas.
Note that the above definition doesn't have $f_i $ to be a diffeomorphism hence I don't have an isomorphism $df_p :T_pM \to T_{f_i(p)}\Bbb R^n.$
If you are really talking about a chart map only, then the answer to your question is yes. The point here is that in the definition of a manifold, you just require the chart maps to be homomorphisms to open subsets in $\mathbb R^n$. But once you have chosen an atlas and thus defined a smooth structure on $M$, you can see that the chart maps of your atlas are indeed smooth. More generally, the charts in the maximal compatible atlas are exactly the diffeomorphisms between open subsets of $M$ and open subsets of $\mathbb R^n$. Thus, you can form derivatives and they do induce linear isomorphisms between tangent spaces on $M$ and on $\mathbb R^n$.
The argument for the description of the charts in the maximal atlas goes as follows: Take your fixed atlas $\{(U_i,f_i):i\in I\}$. Then a chart in the maximal atlas is an open subset $U\subset M$ (which means that $U\cap U_i$ is open in $U_i$ for each $i$) together with a homeomorphism $f:U\to V$, where $V$ is open in $\mathbb R^n$ such that the chart $(U,f)$ is compatible with $(U_i,f_i)$ for each $i$. But this means that for each $i$, the maps $f\circ f_i^{-1}:f_i(U\cap U_i)\to V$ and $f_i\circ f^{-1}:f(U\cap U_i)\to f_i(U_i)$ are smooth as maps between open subsets of $\mathbb R^n$. But the former condition exactly says that $f:U\to V$ is smooth while the latter says that $f^{-1}:V\to U$ is smooth, so $f$ is a diffeomorphism.