There is a diffeomorphism $f$ of $M$ such that $f(x_i) = y_i$ and $df_{x_i}(v_i) = w_i$ 
Let $x_1,...,x_k$ and $y_1,...,y_k$ be two sets of distinct points in a connected smooth manifold $M$ with $\dim M>1$, and $v_1,...,v_k$ and $w_1,...,w_k$ be the corresponding two sets of nonzero tangent vectors at these points. Show that there is a diffeomorphism $f$ of $M$ such that $f(x_i) = y_i$ and $df_{x_i}(v_i)= w_i$ for $i =1,2,...,k$.

I think I can prove for the case $k=1$. First suppose $x,y$ are close enough so that there is an open coordinate ball chart $(B,\varphi)$ containing $x$ and $y$. Let $(x^i)$ a coordinate on $B$ and let $v_1 = v^i{\partial\over\partial x^i}\big|_x$ and $w_1 = w^i{\partial\over\partial x^i}\big|_y$. Define a curve $\gamma:[0,1]\to B$ from $x$ to $y$ such that $\gamma'(0) = (v^1,...,v^n)$ and $\gamma'(1) = (w^1,...,w^n)$ in local coordinate. Then this gives a smooth vector field $X$ along $\gamma$ such that $X_x = v_1$ and $X_y =w_1$. By the uniqueness of the flow, the corresponding flow of $X$ is $\gamma$. Since the image of $\gamma$ is closed, we can extend $X$ onto $M$ with compact support (also denoted by $X$). Since $X$ has compact support, $X$ admits the global flow $\Phi$. Then $\Phi_1:M\to M$ is a diffeomorphism such that $\Phi_1(x) =y$ and $\color{red}{d(\Phi_1)_x(v_1) =w_1}$. The usual connectedness argument proves the statement.
I'm not quite confident about my $k=1$ proof (the red part) but it should be like this (I guess). But for $k>1$, I have no any idea. Please help.
The vector field along the curve I have in mind:

 A: Here is a complete and self-contained proof that is long, but is divided into several elementary parts.
Lemma 1

Let $y\in \Bbb R^n$ and $r>\|y\|$.
Then there exists a diffeomorphism $f\colon \Bbb R^n \to \Bbb R^n$ with compact support in $B(0,r)$ and $f(0)=y$.

Proof:
Consider the constant vector field $X = y$, and $\varphi\colon \Bbb R^n\to \Bbb R$ a smooth bump function such that $\varphi\equiv 1$ on $B(0,\frac{r+\|y\|}{2})$ and $\varphi\equiv 0$ on $\Bbb R^n\setminus B(0,r)$.
Let $\tilde{X}=\varphi X$, which is a vector field with compact support in $B(0,r)$, equal to $X$ on $B(0,\frac{r+\|y\|}{2})$.
Let $\tilde{F}_t$ be the flow of $\tilde{X}$.
Then $f=\tilde{F}_1$ is a solution.
$\blacksquare$
Lemma 2

Let $M^n$ be a connected manifold of dimension $n\geqslant 2$.
Let $x\neq y \in M$, and let $\gamma \colon [0,1]\to M$ be a smooth injective path joining $x$ and $y$.
Let $U$ be an open neighbourhood of the path $\gamma$ in $M$.
Then there exists a diffeomorphism $f\colon M \to M$ with compact support in $U$ such that $f(x) = y$.

Proof:
By compactness of $\gamma([0,1])$, cover the path $\gamma$ with a finite number of charts $(U_i,\phi_i)_{0\leqslant i \leqslant l}$ such that

*

*$U_i \subset U$,

*$\phi_i \colon U \to \Bbb R^n$ is a diffeomorphism,

*$x_0=x$, $x_l=y$,

*$\forall i \in \{0,\ldots,l-1\},\quad x_{i+1}\in U_i$,

*$\forall i \in \{0,\ldots,l\},\quad \phi_i(x_i) = 0$.

For $i\in \{0,\ldots,l-1\}$, Lemma 1 provides the existence of a diffeomorphism $f_i\colon \Bbb R^n \to \Bbb R^n$ with compact support such that $f_i(0) = \phi_i(x_{i+1})$.
Let $\tilde{f}_i = \phi_i^{-1}\circ f_i \circ \phi_i$, extended by the identity oustisde of $U_i$.
Then $\tilde{f}_i \colon M \to M$ is a diffeomorphism with compact support in $U$ such that $\tilde{f}_i(x_i) = x_{i+1}$.
It follows that $f = \tilde{f}_{l-1}\circ \cdots \circ \tilde{f}_0$ is a solution.
$\blacksquare$
Lemma 3

Let $v,w\in \Bbb R^n\setminus \{0\}$, $n\geqslant 2$, and $r>0$.
Then there exists a diffeomorphism $f\colon \Bbb R^n \to \Bbb R^n$, with compact support in $B(0,r)$, such that $f(0)=0$ and $d_0f(v)=w$.

Proof:
Let $\lambda =\frac{\|w\|}{\|v\|}>0$.

*

*If $v$ and $w$ are colinear with $w=\lambda v$, define, for $t\in \Bbb R$, $F_t = \lambda^t\mathrm{Id}_{\Bbb R^n}$.

*If $v$ and $w$ are not colinear, let $P$ be the linear plane spanned by $v$ and $w$, oriented such that $\{v,w\}$ is a direct basis.
Let $\theta$ be the oriented angle between $v$ and $w$.
If $w=-\lambda v$, take a third vector which is not colinear to $v$, say $w'$, and consider the plane $P$ spanned by $v,w'$, and let $\theta=\pi$ [here we need $n\geqslant 2$].
For $t\in \Bbb R$, define $R_t$ to be linear isomorphism of $\Bbb R^n$ such that its restriction to $P$ is the rotation of angle $t\theta$, and its restriction to $P^{\perp}$ is the identity map.
Define $F_t = \lambda^t R_t$.

Then $(F_t)$ is a $1$-parameter subgroup of diffeomorphisms.
Let $X$ be its infinitesimal generator.
Let $\varphi \colon \Bbb R^n \to \Bbb R$ be a bump function such that $\varphi\equiv 1$ on $B(0,\frac{r}{2})$ and $\varphi\equiv 0$ on $\Bbb R^n \setminus B(0,r)$.
Define a vector field $\tilde{X}$ by $\tilde{X} = \varphi X$, and consider $(\tilde{F}_t)$ its flow.
Then $f=\tilde{F}_1$ is a solution.
$\blacksquare$
We are now able to prove an slightly different version of your statement, in the case $k=1$.
Proposition 4

Let $M^n$ be a connected manifold of dimension $n\geqslant 2$.
Let $x,y\in M$ and $v\in T_xM$, $w\in T_yM$ be two non-zero tangent vectors.
Let $U$ be a connected open neighbourhood of $\{x,y\}$ in $M$.
Then there exists a diffeomorphism $f\colon M\to M$ with compact support in $U$, such that $f(x)=y$ and $d_xf(v)=w$.

Proof:
Consider first the case $x=y$.
In that case, let $(V,\phi)$ be a chart such that $V\subset U$, $\phi\colon U \to \Bbb R^n$ is a diffeomorphism, and $\phi(x)=0$.
By Lemma 3, there exists a diffeomorphism $\tilde{f}\colon \Bbb R^n \to \Bbb R^n$ such that $\tilde{f}(0)=0$ and $d_0\tilde{f}\left(d_x\phi(v)\right) = d_x\phi(w)$.
Then $f = \phi^{-1}\circ \tilde{f} \circ \phi$, extended by the identity outside of $V$, is a solution.
Now, consider the case $x\neq y$, let $\gamma\colon [0,1]\to M$ be a smooth path whose support lies in $U$.
By Lemma 2, there exists a diffeomorphism $f_1\colon M\to M$ with compact support in $U$ sending $x$ to $y$.
Let $w' = d_x{f_1}(v) \in T_yM$.
By the first case, there exists a diffeomorphism $f_2\colon M\to M$ with compact support in $U$ such that $f_2(y)=y$ and $d_y{f_2}(w')=w$.
Then $f = f_2\circ f_1$ is a solution.
$\blacksquare$
We can now prove the result you want to show.
Theorem

Let $M^n$ be a connected manifold of dimension $n\geqslant 2$.
Let $\{x_1,\ldots,x_k\}$ and $\{y_1,\ldots,y_k\}$ be two sets of $k$ distincts points in $M$, $k\geqslant 1$.
For all $i\in \{1,\ldots, k\}$, let $v_i\in T_{x_i}M\setminus \{0\}$ and $w_i\in T_{y_i}M\setminus \{0\}$.
Then there exists a diffeomorphism $f\colon M\to M$ such that for each $i\in \{1,\ldots,k\}$, $f(x_i) = y_i$ and $d_{x_i}f(v_i) = w_i$.

Proof:
Let $U_1,\ldots U_k$ be $k$ pairwise disjoint connected open subsets with $x_i,y_i \in U_i$ [see Lemma 5 below].
For each $i$, Proposition 4 yields the existence of a diffeomorphism $f_i\colon M\to M$ with compact support in $U_i$, such that $f_i(x_i) = y_i$ and $d_{x_i}f(v_i) = w_i$.
Then $f = f_1\circ \cdots \circ f_k$ is a solution.
$\blacksquare$
Remark:
We have in fact shown the following more precise result:

Let $M^n$ be a connected manifold of dimension $n\geqslant 2$.
For any $k\geqslant 1$, the group of diffeomorphisms with compact support that are isotopic to the identity acts $k$-transitively on the complement of the zero section in the tangent bundle of $M$.


For the sake of completeness, let us show the following Lemma:
Lemma 5

Let $M^n$ be a connected manifold of dimension $n\geqslant 2$.
Let $x_1,\ldots, x_k$ and $y_1,\ldots,y_k$ be two sets of distinct points.
Then there exists contractible open subsets $U_1,\ldots,U_k$ with $x_i,y_i\in U_i$, such that their closures contractible and are pairwise disjoint.

Proof:
By induction:

*

*For $k=1$: if $x_1=y_1$, take a small contractible neighbourhood of $x_1$ for $U_1$.
If $x_1\neq y_1$, take any small tubular neighbourhood of an injective path joining $x_1$ and $y_1$ for $U_1$.


*Assume $x_1,\ldots,x_{k+1}$, $y_1,\ldots,y_{k+1}$ are as in the statement of the Lemma.
By induction hypothesis, let $U_1,\ldots,U_k$ be disjoint contractible open subsets such that $x_i,y_i\in U_i$ for $i=\{1,\ldots,k\}$, and whose closures are pairwise disjoint.
Take them small enough such that $x_{k+1},y_{k+1}\notin U_i$ for any $i$.
If $x_{k+1}=y_{k+1}$, any small contractible neighbourhood of $x_{k+1}$ works for $U_{k+1}$.
Assume then that $x_{k+1}\neq y_{k+1}$.
Since $\bar{U_1},\ldots,\bar{U_k}$ are contractible, $M\setminus\left(\bar{U_1}\cup \ldots \cup \bar{U_k}\right)$ is homotopy equivalent to $M\setminus \{k \text{ points}\}$.
In particular, it is path-connected.
Since $\tilde{M}=M\setminus \left(\bar{U_1}\cup\ldots \bar{U_k}\right)$ is an open subset of $M$, it is then a connected manifold of dimension $n$, with $x_{k+1},y_{k+1}\in \tilde{M}$.
Apply the $k=1$ case to $\tilde{M},x_{k+1},y_{k+1}$ in order to conclude the proof.
