Let $M$ be $C^{\infty}$manifold and $x\in M$. We define $(k,r)$velocity space at x as
$(T_k^rM)_x:=J_0^{r}(\mathbb{R}^k,M)_x$.Can we define vector space structure on $(T_k^rM)_x$?



Nice answer BS. I was about to post something similar but I didn't have a proof of the nonlinearity of the action of $k$jets of diffeomorphisms. One additional remark that might help the OP is that you can put a vector space structure on $J^k_0({\mathbb R}^n, M)_x$ if you have additional structure on $M$. Basically you need to be able to determine a family of coordinates (actually $k$jets of coordinates) that are related by linear transformations to avoid the nonlinear action of diffeomorphisms when you change coordinates. Also you want to choose different coordinates at each point of $M$. Sufficient would be to choose at each $x \in M$ the $k$jet of a diffeomorphism from $M$ to $T_xM$ sending $x$ to $0 \in T_xM$. For example if $M$ is Riemannian the $k$jet of the inverse of the exponential map would do or if $M$ is a submanifold orthogonal projection onto the tangent subspace would work. In such a case composition with the chosen $k$jet of a diffeomorphism defines a bijection $$ J^k_0({\mathbb R}^n, M)_x \to J^k_0({\mathbb R}^n, T_xM)_0 $$ and the latter space is a vector space because $T_xM$ is a vector space. 


I assume you ask for a "natural" vector space structure, one for which the natural group action of $r$jets of diffeomorphisms of $M$ fixing $x$ is linear, and that the vector structure is smooth in standard coordinates. Then the answer is negative for all $r>1$ (for $r=1$ it is of course positive, the space is $(T_x M)^k$). For the proof, let $\phi:(\mathbb{R}^n,0)\to(\mathbb{R}^n,0)$ be a germ of diffeomorphism of the form $\phi(x)=x+h(x)$ with $h\neq 0$ homogeneous of degree $r$, and $f:(\mathbb{R}^k,0)\to(\mathbb{R}^n,0)$. Then for $k\lt r$, $$D^k(\phi\circ f)(0)=D^kf(0),$$ and $$D^r(\phi\circ f)(0)=D^rf(0)+D^rh(Df(0),...,Df(0))$$. Hence the action of $J_0^r\phi$ on $J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0$ is tangent to the identity at the jet of the zero map $\mathbf{0}$, but is not the identity. Now assume for contradiction that there is a "natural" vector space structure. Since the $\mathbf{0}$ is the only jet fixed by all $r$jets of diffeomorphisms fixing $0$, it is the only possible zero of a natural vector space structure. Now, by assumption the action of $J_0^r\phi$ is linearizable by a smooth ($C^1$ is enough) diffeomorphism $$\psi: (J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0,\mathbf{0}) \to (\mathbb{R}^N,0),$$ but the only possible candidate for the linear map $\psi\circ J_0^r\phi\circ\psi^{1}$ from $(\mathbb{R}^N,0)$ to itself is its derivative at 0, i.e. the identity, a contradiction. PS: the natural structure on jets is a bit complicated. There is a sequence of fiber bundles $$T_k^r M\to T^{r1}_k M\to...\to T^1_k M\to M,$$ where the rightmost map is a vector bundle, but the others are affine bundles under the vector bundles $P_k^r M \simeq S^r(\mathbb{R}^k)^*\otimes TM$ of jets in $T_k^r M$ which map to constant jets in $T^{r1}_k M$. To see that the $P_k^r M$ are vector bundles, one may observe that a $r$jet of diffeomorphism $J_x^r\phi$ of $M$ acts on $(P_k^r M)_x$ only through $J_x^1\phi$. It is remarkable that the choice of an affine connection on $M$, which may be viewed as a splitting of $$0\to P_1^2 M\to T_1^2 M\to T^1_1M=TM\to 0,$$ is sufficient to put a vector bundle structure on all jet bundles, by using covariant derivatives, see this wikipedia article and its references for more details, notably on the important Cartan distribution on jet bundles. 

