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
I suppose you are talking about deRham cohomology. Then it would be wise to take a look at the work of Richard Melrose, e.g. his book The Atiyah-Patodi-Singer Index Theorem.
On page 65 he discusses deRham cohomology for manifolds with boundary (which can be easily generalized to the corner case, as was also done by him!). On manifolds with corners something interesting happens: there are different versions of reasonable vector fields (and—by duality—differential forms), e.g.
(It can be shown that $d$ preserves the classes 1.-3., giving a deRham complex whose cohomology can be computed) Melrose points out (compact with boundary case) that in cases 1. and 3. the deRham cohomology is canonically isomorphic to the singular cohomology of the underlying topological space. For the 2nd case also the cohomology of the boundary enters (via a degree shift).
I should also point out that there is also a working Morse theory on manifolds with corners, see for example
and [this broken link EDIT perhaps someone else can extract a result -DR]
Furthermore it is easy to construct "invariants" of the manifold with corners by also taking into account its corners (but be careful with respect to which transformations this is an invariant)!