General relativity: Why don't these two differentials commute?

1. 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.

2. 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!