I encountered some metric today defined by $$ ds^2 = - \left( 1 - \frac{2GM}{r} \right) dv^2 + (dv dr + dr dv) + r^2 d \Omega^2 $$
In all education I've done until now $$dv dr = dr dv.$$ Why is this no longer the case? I suspect this has something to do with tensors, but I am not sure why.
More generally, a metric tensor $$\mathbb{g} ~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}M)\right)$$ is a section in the symmetric tensor product $${\rm Sym}^2(T^{\ast}M)~=~T^{\ast}M\odot T^{\ast}M$$ over the cotangent bundle $T^{\ast}M$. In other words, $\mathbb{g}$ is a symmetric $(0,2)$ covariant tensor field.
In a coordinate chart $U\subseteq M$, it takes the form $$\mathbb{g}|_U ~=~ g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu},$$ with the manifest rule $$ \mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}~=~\mathrm{d}x^{\nu}\odot \mathrm{d}x^{\mu}, $$ cf. OP's question.
That is the Eddington-Finkelstein metric (one of the forms) and it can indeed be written as:
$$ ds^2 = - \left( 1 - \frac{2GM}{r} \right) dv^2 + 2dv dr + r^2 d \Omega^2 $$
Why your book doesn't write it that way I don't know - you'd have to ask the author.
This is the Eddington-Finkelstein metric for a Schwarzschild geometry, and the differentials do commute. I believe they put the line element in this form in order to highlight the that the components of the metric are
$$g_{vv}=\left(1-\frac{2GM}{r}\right),\hspace{0.5cm}\,g_{vr}=g_{rv}=-1$$ $$g_{\theta\theta}=-r^2,\hspace{0.5cm}g_{\phi\phi}=-\sin^2{\theta}$$
(Note that I'm using a $(+,-,-,-)$ signature.) If this is an introductory text, then the author may have done this to show that off-diagonal elements of the metric have to be halved when being read off from the line element.
I hope this helps!
Another way of thinking about this:
In general, tensors with "down" indices are maps from the space of vectors${}^{1}$ to the space of functions. So, a tensor $\bf T$ is a function that takes two vectors and spits out a real number, and you can think of it in the "index-free" manner as:
$${\bf T}\left({\vec v_{1}}, {\vec v_{2}}\right) = f(x^{a})$$
There is also a further restriction on $T$: it is multilinear in both of its coordinates. So, if you multiply either vector by a scalar, you multiply $f$ by the same scalar; and if you replace either vector by a sum of two other vectors, the result is the sum of the tensor acting on the individual vectors. This forces $T$ to be representable as a matrix, which leads us to the familiar notation:
$$f(x^{c}) = v_{1}^{a}T_{ab}v_{2}^{b}$$
Now, it should be obvious that ${\bf T}$ can, in principle, act differently on $v_{1}$ and $v_{2}$, and this happens if $T_{ab}$ is not symmetric.
Now, what does this have to do with differentials? Well, remember that we can express the basis of the vector space with partial derivatives, so that:
$${\vec v} = v^{a}\frac{\partial}{\partial x^{a}}$$
which makes $\vec v$ acting on a function $f$ the directional derivative of $f$ along $\vec v$. So, partial derivatives are the basis of our "up" space of vectors. What is the basis of the "down" space of vectors? Well, we need something that has an index, let's call it $e^{a}$, we also need:
$$\frac{\partial}{\partial x^{a}}e^{b} = \delta^{b}{}_{a}$$
Well, if we choose $e^{a} = dx^{a}$, this should be obviously true. So, just like we can express:
$${\vec v} = v^{a}\frac{\partial}{\partial x^{a}}$$
we can also express:
$${\bf T} = T_{ab}dx^{a}dx^{b}$$
And the only way we can also have a $\bf T$ that acts differently on its first and second argument is if we also have $dx^{a}dx^{b} \neq dx^{b}dx^{a}$ when $a \neq b$
Finally, note that a lot of this is moot, because metric tensors are defined to be symmetric: the basal requirement for a dot product is that ${\vec v} \cdot {\vec w} = {\vec w} \cdot {\vec v} = g_{ab}v^{a}w^{b} = g_{ab}w^{a}v^{b}$
${}^{1}$ Yes, I know we're talking about vector fields and tensor fields, rather than vectors and tensors, but let's not complicate this by making that distinction right now.
Written out explicitly (see this Wikipedia article):
$$v=t+r^{*}$$
$$dr^{*}=(1-{\frac {2GM} r})^{-1}dr$$
so
$$dv=dt+dr^{*}=dt+(1-{\frac {2GM} r})^{-1}dr$$
Then
$$dvdr=(dt+(1-{\frac{2GM} r})^{-1}dr)dr=dtdr+(1-{\frac{2GM}{r}})^{-1}d^2r$$
and$$drdv=dr(dt+(1-{\frac{2GM} r})^{-1}dr)=drdt+(1-{\frac{2GM} r})^{-1}dr^2$$
wich are equal because the infinitesimal increments $dr$ and $dt$ are infinitesimal increments in numbers and $r$ is a number too.
So $$dvdr=drdv$$
which is why the metric in the article has a $2dvdr$ term.
In multilinear algebra, there are several ways to construct new vector spaces from old ones: for a given (real, finite dimensional) vector space $V$, we can construct its dual $V^\ast$, given vector spaces $V$ and $W$ we can construct $V\otimes W$, and we can construct $V\odot V$, $V\wedge V$ (and many more of course). Also these constructions can be repeated.
Typically the latter, $V\odot V$, $V\wedge V$, are constructed by mathematicians as quotients of $V\otimes V$, which works much more generally than for finite dimensional vector spaces over a field. In our special case however (over a field), they can also be constructed as subspaces of $V\otimes V$, in the symmetric case the subspace generated by elements of the form $v\otimes v' + v'\otimes v$, and the alternating case by elements of the form $v\otimes v' - v'\otimes v$.
Physicists, by historical accident, by the lack of need for more generality or by the emphasis on calculation, focus on the second approach. If you make use of several natural isomorphisms like $V^{\ast\ast} \cong V$, $\text{Hom}(V,V) \equiv V^\ast\otimes V$, you get quite far in describing all kinds of derived vector spaces purely in terms of tensor products and duals. When $V = T_xM$, the tangent space of a manifold, the elements of the resulting objects are called tensors. One ugly artefact of this approach that I am aware of, is that the one-dimensional top-exterior power $\Lambda^nV$ of an $n$-dimensional vector space $V$ is viewed as something that is almost like a scalar, but not quite; its elements are called tensor densities.
Another approach popular under physicists is to consider the new vector spaces as multilinear maps $V\times\cdots V\times V^\ast\times\cdots\times V^\ast\to\Bbb R$. If you allow restrictions on these maps, like symmetry or antisymmetry in some of the arguments, and you make use of $V\otimes W \cong W\otimes V$, you can describe most tensors of interests to physicists.
Finally, a metric tensor is a section $g$ of $T^\ast M\odot T^\ast M$, like Qmechanic said. Here $T^\ast M$ can be seen as the union of all $T_x^\ast M$ (along with a given differentiable structure), and a section assigns an element $g(x)\in T_x^\ast M\odot T_x^\ast M$ to each $x\in M$. Depending on your explicit construction, you view elements of $T_x^\ast M\odot T_x^\ast M$ as being generated by $\alpha\otimes\beta + \beta\otimes\alpha$, for $\alpha,\beta\in T_x^\ast M$, i.e. $drdv + dvdr$ in your case, or as products $\alpha\odot\beta$ or $\alpha\beta$ with the understanding that $\alpha\odot\beta = \beta\odot\alpha$ and $\alpha\beta = \beta\alpha$.
