It's not very clear to me which properties you expect from that object, but interpreting the question literally the answer is no: there is no natural way to associate to some vectors $v_1,v_2,\ldots,v_k\in T_x M$ and a function $f\in C^\infty(M)$ an object corresponding "only" to the partial derivative $\frac{\partial ^kf}{\partial v_1\cdots\partial v_k}$ $k\geq 2$. As Otis remarks there are jets, but these encode all lower order derivatives too, and there is no natural way to disentangle them. The reason is that all functions with $df(x)\neq 0$ look locally the same, so they would all have "second derivative" zero. This is what Will says in his remark.

On the other hand, if the function $f$ vanishes to order $k-1$ in $x$ (in terms of jets: makes $k-1$-contact with a constant function), then the partial derivative depends only on the vectors $v_1,v_2,\ldots,v_k\in T_x M$ and is an element of $T_{f(x)}Y$.

If you assume a metric on $M$ you should also be able to save the situation, maybe extending the vectors to geoedesics and then by parallel transport to vector fields... but I'm not sure about that.

**Edit in response to the comment**: a high level answer is this: the fibers of the projection from the space of jets of order k to jets of order k-1 are naturally affine spaces over the vector space $S^k(T^*_x M)\otimes T_y N$. So if two maps have contact of order k-1, then their jets of order k differ by a k-symmetric map $T_x M\times \cdots T_x M \to T_y N$, which in your case is the map you expect in local coordinates $v_1,\ldots ,v_k\to \frac{\partial ^kf}{\partial v_1\cdots\partial v_k}$. There are different proofs for this affine structure in the literature on jets (personally I find non of them particularly enlightening).

In the case that $N=\mathbb{R}$ it is not so hard to show directly, that if you extend your vectors to vector fields $X_1,\ldots,X_2$ then the map $v_1,\ldots ,v_k\mapsto X_k(X_{k-1}(\cdots (X_1(f))\cdots)(x)$ is independent of the choice of extensions when $f$ vanishes to order k-1 in $x$.

isa big deal, as the construction will depend on the metric and not be intrinsic to the differentiable manifold. If you accept the object $\partial^k f/\partial \bar v$ you seek to depend upon a choice of metric, you should revise your question accordingly. – Benoît Kloeckner Jul 15 '12 at 12:15