"Higher" tangent space Let $k[\varepsilon] = k[t]/t^2$ be the algebra of dual numbers i.e. $\varepsilon^2 = 0$, here $k$ is a field. Then for a scheme $X$ over $k$ and $x \in X(k)$ we may consider the set $X(k[\varepsilon])_x$ of $k[\varepsilon]$-valued points supported at $x$ i.e. this set is the preimage of $x$ via the mapping $X(k[\varepsilon]) \to X(k)$ induced by the canonical projection $k[\varepsilon] \to k$, $\varepsilon \mapsto 0$.
One interesting fact is, that $X(k[\varepsilon])_x$ has a natural structure of a $k$-vector space. (Indeed, a point in this set defines a local homomorphism $\varphi: \mathcal O_{X,x} \to k[\varepsilon]$ which we may write as $\varphi(s) = s(x) + \dot\varphi(s)\varepsilon$ for a germ $s$. For two such morphisms $\varphi, \varphi'$ and $\alpha \in k$ we define $\varphi + \alpha \varphi'$ as the homomorphism $s \mapsto s(x) + (\dot\varphi(s) + \alpha\dot\varphi'(s))\varepsilon$). This is just a different description of the Zariski-tangent space of $X$ at $x$.
Q: Now, what happens if I replace $k[\varepsilon]$ by $A_m = k[t]/(t^{m+1})$, m > 1? Do we also have a "natural" vector space structure on $X(A_m)_x$?
 A: I do not think so. Consider the following example.
Let $X = \operatorname{Spec}(k[x,y]/(xy))\subseteq\mathbb A^2.$ Suppose $j:\operatorname{Spec}(k[t]/(t^3))\to X$ maps the unique point to the origin. This map is determined by $j^*:k[x,y]/(xy)\to k[t]/(t^3)$ which in turn is determined by $$j^*(x) = a_0+a_1t+a_2t^2, j^*(y) = b_0+b_1t+b_2t^2$$ satisfying 
$$0=j^*(x)j^*(y) = a_0b_0+(a_0b_1+a_1b_0)t+(a_0b_2+a_1b_1+a_2b_0)t^2.$$ In particular, each of these coefficients must be zero. Treating the coefficients as coordinates in a larger space, the second jet scheme of $X$ is given by their vanishing
$$J_2(X) = \operatorname{Spec}(k[a_0,a_1,a_2,b_0,b_1,b_2]/(a_0b_0,a_0b_1+a_1b_0,a_0b_2+a_1b_1+a_2b_0)).$$
(A jet scheme is a higher order version of the tangent bundle. So we're interested in the fibres of this space.) Now since we've chosen the jet at the origin, we also have $a_0,b_0 = 0.$ That is, the fibre of the second jet scheme of $X$ over the origin is 
$$J_2(X)_{(0,0)} = \operatorname{Spec}(k[a_1,a_2,b_1,b_2]/(a_1b_1)).$$
But now we observe that $J_2(X)_{(0,0)}$ has two components: $(a_1=0)$ and $(b_1=0)$, both of which are isomorphic to $\mathbb A^3.$ I believe this answers the question, somewhat indirectly: what natural vector space structure does this space have? It seems to me that there is none.
A possibly relevant observation is that in general $J_1(X) = TX = \mathcal{Spec}(\operatorname{Sym}\Omega_{X/k})$, i.e., the tangent bundle can be realised as the (global) spectrum of the symmetric algebra on the cotangent sheaf. On the other hand, for higher jet schemes there is no equivalent module whose symmetric power gives the regular functions (although one can work with algebras of Hasse-Schmidt derivations instead). 
