function of class C ^ 1 on manifolds Let $M$ be a differentiable manifold with finite dimension $ m $. 
Let $ f:M\rightarrow M $ a function of class $C^1$. I have a doubt about what this implies (1) or (2):


*

*$x \in M \rightarrow D_xf \in \mathcal{L}(T_xM, T_{f(x)}M)$ is continuous    

*$Df:TM \rightarrow TM$ is continuous
Note that (1) is the natural extension of $f:\mathbb{R}^m \rightarrow \mathbb{R}^m $, but the set of arrival would not be well defined (varies with $x$ ). On the other hand, (2) does not imply (in my opinion) that $M \rightarrow D_xf $ is continuous.
Would appreciate any suggestions on the proper way to interpret this concept 
 A: In general, given smooth manifolds $M$ and $N$ and a map $f: M\to N$. $f$ is called a map of class $C^k$ if for all $x\in M$, there are smooth charts $(U , \phi)$ of $M$, $(V, \psi)$ of $N$, so that $x\in U$ and $f(U) \subset V$ and the composition 
$$F= \psi \circ f \circ \phi^{-1} : \phi(U) \to \psi(V)$$
is $C^k$ in the usual sense (between open sets in Euclidean space). Thus, if $f$ is $C^k$ for $k\geq 1$, then so is $F$ and thus $dF$ is of class $C^{k-1}$, meaning that all the entries of $dF$ are functions of class $C^{k-1}$. 
First we show the following (which relate the class of $f$ with that of $Df$)
Claim 1: Let $f: M \to N$ be a $C^1$-map. Then $f$ is of class $C^k$, $k\geq 1$ if and only if $Df: TM \to TN$ is of class $C^{k-1}$. 
Proof of Claim 1: Note that if $(U, \phi)$,are chart on $M$, then one can define charts $(TU , \tilde \phi)$ on $TU \subset TM$ given by 
$$\tilde \phi [\gamma ] = (x, (\phi \circ \gamma)'(0)) \subset \phi(U) \times \mathbb R^m$$
for all $[\gamma] \in T_xM$.  
The map $Df: TM \to TN$ is defined by $[\gamma] \mapsto [f\circ \gamma]$. Thus under the chart $(TU, \tilde \phi)$ and $(TV, \tilde \psi)$, 
$$(\tilde \psi \circ Df \circ \tilde \phi^{-1} )(p, v) = (\tilde \psi \circ Df) [\gamma],$$
where $[\gamma]\in T_xM$ such that $\tilde \phi [\gamma] = (p, v)$ (so $\phi(x) = p)$. For example, we take $\gamma(t) = \phi^{-1}(p+ tv)$. Then
$$(\tilde \psi \circ Df) [\gamma] = \tilde \psi [f\circ \gamma] = (F(x), (\psi \circ f\circ \gamma)'(0))$$
As we have
$$(\psi \circ f\circ \gamma)'(0) = \frac{d}{dt}\bigg|_{t=0} \psi \circ f \circ \phi^{-1} (x + tv) = \frac{d}{dt}\bigg|_{t=0} F(p+ tv) = dF_p (v)$$
by the usual Chain rule in the last step. So 
$$(\tilde \psi \circ Df \circ \tilde \phi^{-1} )(p, v) = (F(p) , dF_p(v)).$$
That is, $Df$ is $C^{k-1} \Leftrightarrow p\mapsto dF_p$ is $C^{k-1} \Leftrightarrow F$ is $C^k \Leftrightarrow f$ is $C^k$. Thus Claim 1 is shown.
(Note in particular, $f:M \to M$ is $C^1$ if and only if $Df: TM \to TM$ is continuous ($C^0$).)
Next, we deal with question (1). It is rather long if I check all the smoothness conditions, so I will just be brief here. You want to show that $x\mapsto (D_x f: T_xM \to T_{f(x)}N)$ is continuous. But you have to specify a domain for this map. First we define a smooth bundle $f^*TN$ on $M$ given by 
$$f^*TN = \{ (x, v) \in M \times TN: f(x) = \pi_N (v)\}.$$
Next we define the vector bundle $End(TM , f^*TN)$ on $M$, so that at $x\in M$, the fiber $End(TM, f^*TN)_x$ is the linear space 
$$End(T_xM, (f^*TN)_x) = End(T_xM, T_f(x)N)$$
(or $\mathcal L(T_xM , T_{f(x)}N)$ using your notation). Thus the map $\tilde Df : M\to End(TM, f^*TN)$ given by $x\mapsto D_xf$ is a section. 
Claim 2: Let $f:M \to N$ be a $C^1$ map. Then $f$ is of class $C^k$, $k\geq 2$, if and only if $\tilde D f$ is a $C^{k-1}$ map. 
Proof of claim 2: Using the same notations, under a chart $(U, \phi)$ of $M$, we want to specify a chart $(End(TM, f^*TN)|_U, \hat \Phi)$ on $D_x f \in End(TM, f^*TN)$. Now for any $x\in M$ and $(U, \phi)$ a chart containing $x$, define the isomorphism $\hat \phi_x : T_x M \to \mathbb R^m$ by 
$$\hat\phi_x [\gamma] = (\phi\circ \gamma)'(0). $$
Then we can define 
$$\hat\Phi : End(TM, f^*TN)|_U \to \mathbb R^m \times End(\mathbb R^m , \mathbb R^n)$$
by $\hat\Phi (A_x) = (\phi(x), \hat\psi_{f(x)}\circ A_x \circ \hat\phi_x^{-1})$ for all $A_x : T_xM \to T_{f(x)}N$.
Under this chart, we have (write $\phi(x) = p$)
$$ \hat\Phi\circ \tilde D f \circ \phi^{-1} (p)= \hat\Phi D_xf = (p, dF_p)$$
Thus $f$ is of class $C^k$ if and only if $\tilde D f$ is of class $C^{k-1}$. Thus Claim 2 is shown. 
That is, both (1) and (2) are equivalent to the statement that $f$ is $C^1$. 
