The existence of flows in the direction of a vector field seems to require Hausdorff; indeed, consider the vector field $\frac{\partial}{\partial x}$ on the line-with-two-origins. We have no global existence of a flow for any positive t, even if we make our space compact (that is, considering the circle-with-one-point-doubled). If the nonexistence of the flow is not visibly clear, consider instead the real line with the interval [0,1] doubled.
Also, partitions of unity do not exist; for example, in the line with two origins, take the open cover by "the line plus the first origin" and "the line plus the second origin". There is no partition of unity subordinate to this cover (the values at each origin would have to be 1).
For me, a basic example of the beauty of this function-theoretic approach is the definition of a vector field as a derivation $D\colon C^\infty(M)\to C^\infty(M)$. The proof that such a derivation defines a vector field hinges upon the fact that $Df$ near a point p only depends on $f$ near the point p. To prove this fact you use the fineness of your sheaf $\mathcal{O}_X$, i.e. the existence of partitions of unity. (It is true though that the failure of fineness in the non-Hausdorff case is of a different sort and might not break this particular theorem.) I feel that the existence of partitions of unity, and the implications thereof, is one of the basic fundamentals of approaching smooth manifolds through their functions; more importantly, a good handle on how partitions of unity are used is important to understand the differences that arise when the same approach is extended to more rigid functions (holomorphic, algebraic, etc.).
Now that the question has been edited to ask specifically about Stokes' theorem, let me say a bit more. Stokes' theorem will be false for non-Hausdorff manifolds, because you can (loosely speaking) quotient out by part of your manifold, and thus part of its homology, without killing all of it.
For the simplest example, consider dimension 1, where Stokes' theorem is the fundamental theorem of calculus. Let $X$ be the forked line, the 1-dimensional (non-Hausdorff) manifold which is the real line with the half-ray $[0,\infty)$ doubled. For nonnegative $x$, denote the two copies of $x$ by $x^\bullet$ and $x_\bullet$, and consider the submanifold $M$ consisting of $[-1,0) \cup [0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. The boundary of $M$ consists of the three points $[-1]$ (with negative orientation), $[1^\bullet]$ (with positive orientation), and $[1_\bullet]$ (with positive orientation); to see this, just note that every other point is a manifold point.
Consider the real-valued function on $X$ given by "$f(x)=x$" (by which I mean $f(x^\bullet)=f(x_\bullet)=x$). Its differential is the 1-form which we would naturally call $dx$. Now consider $\int_M dx$; it seems clear that this integral is 3, but I don't actually need this. Stokes' theorem would say that
$\int_M dx=\int_M df = \int_{\partial M}f=f(1^\bullet)+f(1_\bullet)-f(-1)=1+1-(-1)=3$.
This is all fine so far, but now consider the function given by $g(x)=x+10$. Since $dg=dx$, we should have
$\int_M dx=\int_M dg=\int_{\partial M}g=g(1^\bullet)+g(1_\bullet)-g(-1)=11+11-9=13$. Contradiction.
It's possible to explain this by the nonexistence of flows (instead of $df$, consider the flux of the flow by $\nabla f$). But also note that Stokes' theorem, i.e. homology theory, is founded on a well-defined boundary operation. However, without the Hausdorff condition, open submanifolds do not have unique boundaries, as for example $[-1,0)$ inside $X$, and so we can't break up our manifolds into smaller pieces. We can pass to the Hausdorff-ization as Andrew suggests by identifying $0^\bullet$ with $0_\bullet$, but now we lose additivity. Recall that $M$ was the disjoint union of $A=[-1,0)$ and $B=[0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. So in the quotient $\partial [A] = [0]-[-1]$ and $\partial [B] = [1^\bullet]-[0]+[1_\bullet]-[0]=[1^\bullet]+[1_\bullet]-2[0]$, which shows that $\partial [M]\neq \partial [A]+\partial [B]$. This is inconsistent with any sort of Stokes formalism.
Finally, I'd like to point out that Stokes' theorem aside, even rather nice non-Hausdorff manifolds can be significantly more complicated than we might want to deal with. One nice example is the leaf-space of the foliation of the punctured plane by the level sets of the function $f(x,y)=xy$. The leaf-space looks like the union of the lines $y=x$ and $y=-x$, except that the intersection has been blown up to four points, each of which is dense in this subset. In general, any finite graph can be modeled as a non-Hausdorff 1-manifold by blowing up the vertices, and in higher dimensions the situation is even more confusing. So for any introductory explanation, I would strongly recommend requiring Hausdorff until the students have a lot more intuition about manifolds.
Best Answer
The definition that Martin mentions comes close to the definition of a tangent vector which I learnt as an undergraduate.
"Definition: A geometric tangent vector is an equivalence class of germs of smooth maps $\mathbb{R} \to M$ at $0$, where two germs are equivalent if their first order jets at $0$ agree."
More generally, we could say that for any test manifold $U$, a smooth map $U \to TM$ is an equivalence class of germs of maps $U \times \mathbb{R} \to M$ around $U \times \{0\}$, where two germs are equivalent if their first order jets in $\mathbb{R}$-direction are equal.
To make sense out of the equivalence relation, we need, horribile dictu, a small computation in local coordinates with the chain rule. It takes some work to show that the functor $U \mapsto \{ \text{germs of maps} U \times \mathbb{R} \to M\}$ is representable by a vector bundle $TM$. Likewise, one needs some local computations and arguments with charts for that.
The problem of course is that there does not exist a manifold $\ast[\epsilon]$ such that maps $\ast[\epsilon] \to M$ correspond to tangent vectors of $M$. So the functor $M \mapsto TM$ does not have an adjoint. It might be a good idea to use supermanifolds instead. Remeber that there is a $(0,1)$-dimensional supermanifold with function algebra $\mathbb{R} [\epsilon]/ \epsilon^2$. However, I am not expert enough to tell you something more specific about that.
I am deeply convinced that it is indeed necessary to invoke coordinates in some form and that the ''nasty'' computations with the chain rule and the fundamental theorem of calculus are absolutely crucial. If we wish to define a structure in which we can talk about derivatives of maps between manifolds, the most basic properties of the derivative of maps between euclidean spaces ought to play a role. One can of course use mathematical high-tech weaponry of all sorts to hide the local charts carefully. I will not try to do so. Anyway, if you wish to create the tangent bundle not as an object in a functor category, but as a manifold, you need to check that it is a manifold, i.e. you need charts at some point.
The construction of the tangent bundle immediately generalizes to a construction of the frame bundle $Fr(M)$ instead. Because it is so funny, we immediately define the higher frame bundles $Fr^k(M)$.
Definition: a $k$-frame is an equivalence class of germs of local diffeomorphisms $\mathbb{R}^n \to M$, where two germs are equivalent if their $k$-th order jets coincide. More generally, for a manifold $U$, let $Fr^k M (U)$ be the set of germs of smooth maps $U \times \mathbb{R}^n \to M$, which are diffeomorphic in $\mathbb{R}^n $-direction. Equivalence is defined by equality of $k$-jets."
Again, the chain rule is needed to justify the definition of the equivalence relation. Now we define the jet groups $J^k (n)$. Consider the power series ring $R:=\mathbb{R}[[x_1,\ldots ,x_n]]$ and let $G$ be the group of ring automorphisms. By the action on Taylor expansions, we get a homomorphism $Diff(\mathbb{R};0) \to G$. Let $I \subset R$ be the unique maximal ideal. Clearly, the group $G$ preserves the filtration $R \supset I \supset I^2 \supset \ldots$. Hence it acts on the finite-dimensional vector spaces $R / I^{k+1}$. The $k$th jet group $J^k (n)$ is the quotient of $G$ by the kernel of its action on $R / I^{k+1}$. This is easily seen to be a linear algebraic group. There are obvious maps $J^{k+1} (n)\to J^k (n)$ and an equally obvious isomorphism $J^1 (n) \cong GL_n (\mathbb{R})$. The extensions have nilpotent kernel and they are split (take derivatives of polynomial diffeomorphisms), but the splitting is not natural (it is natural with respect to linear maps $\mathbb{R}^n \to \mathbb{R}^n$, but not more generally).
It is easy to see that $U \mapsto Fr^k (M)(U)$ is a torsor over the group $map (U; J^k (n))$ (compose with germs of diffeomorphisms).
The functor $U \mapsto Fr^k (M)(U)$ is representable by a $J^k (n)$-principal bundle. This is done in charts and then by gluing. Again, the functor $Fr^k (M)$ certainly does not have an adjoint.
The second link that Martin gave alludes to an axiomatic characterization of the tangent bundle (and the higher frame bundles as well). It does not quite fit to the functor-of-points ideology, but I think still very useful, because it axiomatizes the gluing and hence I say some words about it as well.
Let $C_n$ be the category of smooth manifolds of dimension $n$ as objects and local diffeomorphisms as morphisms.
A natural fibre bundle on $C_n$ is the following set of data: For each $M \in \operatorname{Ob} (C_n)$, there is a smooth fibre bundle $F_M \to M$ and for each map $f:M \to N$ in $C_n$, there is a bundle map $F_M \to F_N$; plus some obvious functoriality properties. There is also a canonical notion of a morphism of natural fibre bundles. Let $F(0)$ be the fibre of $F_{\mathbb{R}^n} \to \mathbb{R}^n$ at $0$. There is an action of $Diff(\mathbb{R}^n;0)$ (diffeomorphisms fixing the origin) on $F(0)$. Here is an axiomatic characterization of the tangent bundle:
''THEOREM: The tangent bundle is the unique natural fibre bundle on $C_n$, such that $F(0)=\mathbb{R}^n$, with the action of $Diff(\mathbb{R}^n,0)$ given by the first derivative $Diff(\mathbb{R}^n,0) \to GL_n (\mathbb{R})$.''
It is clear that the tangent bundle satisfies these properties, and here is a sketch of uniqueness, in other words, that the tangent bundle is determined by these properties. Let $F_M \to M$ be a natural bundle. I show that it is determined, up to natural isomorphism, by the action on $F(0)$.
First restrict to $M = V = \mathbb{R}^n$. We know that there is an action of the diffeomorphism group $Diff(V)$ on $F_V$, covering the action on $V$. Using the translations $T_x (v):= v +x$, we get a trivialization $F_V \times V \times F(0)$. Let $x \in V$, $f \in Diff(V)$. The action $f:F(x) \to F(f(x))$ is given in this trivialization by the action of $T_{-f(x)} \circ f \circ T_x \in Diff(V;0)$ on $F(0)$, which is known by assumption. This argument shows that $F(0)$ plus the action determines $F_V$ completely.
By restriction to open subsets and naturality, the restriction of $F$ to the full subcategory $O_n$ of manifolds diffeomorphic to some open subset of $V$ is uniquely determined.
Let $M$ be an arbitrary manifold and let $U(i)_{i \in I}$ be the maximal atlas. It can be written as a diagram $U : J \to O_n$, for some indexing category $J$ (take the intersection of the charts into account). The diagram has a colimit in $C_n$, namely $M$ (of course not all colimits in $C_n$ exist). Likewise, we get a diagram $j \mapsto F_{U(j)}$, whose colimit is $F$. Thus a natural bundle is completely determined, up to natural isomorphism, by the action of $Diff (V,=)$ on the fibre $F(0)$.
The existence question for general natural fibre bundles is a bit subtler, for several reasons. There is a theorem by Palais and Terng asserting that the action on $F(0)$ automatically has finite order (it factor through $j^k (n)$ for some $k$). Also, you have to take into account the topology of the diffeomorphism group. But for the tangent bundle, these subtleties do not arise: it is known what the tangent bundle of an open subset of $\mathbb{R}^n$ has to be and how diffeomorphism act on that. The colimit procedure then produces the tangent bundle. The higher frame bundles can be constructed in a similar way.