What's bad about left $\mathbb{H}$-modules? Can you give me non-trivial examples of propositions that can be formulated for every left $k$-module, hold whenever $k$ is a field, but do not hold when $k = \mathbb{H}$ or, more generally, need not hold when $k$ is a division ring (thanks, Bruno Stonek!) which is not a field?
I'm asking because in the theory of vector bundles $\mathbb{H}$-bundles are usually considered alongside those over $\mathbb{R}$ and $\mathbb{C}$, and I'd like to know what to watch out for in this particular case.
 A: Linear algebra works pretty much the same over $\mathbb H$ as over any field. Multilinear algebra, on the other hand, breaks down in places.
You can see this already in the fact that when $V$ and $W$ are left $\mathbb H$-modules, the set of $\mathbb H$-linear maps $\hom_{\mathbb H}(V,W)$ is no longer, in any natural way, an $\mathbb H$-module. As Bruno notes, tensor products also break (unless you are willing to consider also right modules, or bimodules—and then it is you who broke!) 
A: EDIT to add: The following previous question is very related: Linear algebra over a division ring vs. over a field
Your "more generally" is a bit deceptive: you're forgetting that $\mathbb{H}$ is a division ring. In the case where you only drop commutativity of the field, such as in this case, a lot of linear algebra still works, as is witnessed by some authors defining "vector space" as "module over a division ring" (for example, Carlos Ivorra in his freely available Algebra book [in Spanish]).
A very prominent example of some linear algebra that doesn't work when you're not over a commutative ring, is the determinant. This is obvious from its definition (as the sum over all the permutations, etc.)
Another example, as suggested by Soarer in the comments, is given by the tensor product. The tensor product of vector spaces over a field has a natural structure of vector space (over the same field). If you drop commutativity, this is not so. The tensor product of two modules over a not necessarily commutative ring does not carry a natural action of the ring, and thus is only an abelian group. If you accept bimodules instead of modules, then you can still give it a module structure, as explained here, in the "Additional structure" section.
I quote Wikipedia in the Division ring article:

Much of linear algebra may be formulated, and remains correct, for (left) modules over division rings instead of vector spaces over fields. Every module over a division ring has a basis; linear maps between finite-dimensional modules over a division ring can be described by matrices, and the Gaussian elimination algorithm remains applicable. Differences between linear algebra over fields and skew fields occur whenever the order of the factors in a product matters. For example, the proof that the column rank of a matrix over a field equals its row rank yields for matrices over division rings only that the left column rank equals its right row rank: it does not make sense to speak about the rank of a matrix over a division ring.

Some linear algebra that does work over division rings is developed in these notes by Paul Garrett.
