In page 66, chap 9 of the book "classical mechanics point particles and relativity" of Walter Greiner, say: "A surface for which a normal vector may be constructed at any point is called orientable." and then say that Mobius strip is not orientable. Since normal vector $$\boldsymbol{n}(u,v)=\frac{\boldsymbol{r}_u\times \boldsymbol{r}_v}{|\boldsymbol{r}_u\times \boldsymbol{r}_v|}$$ (here, $\boldsymbol{r}(u,v)$ is a surface, $\boldsymbol{r}_u=\frac{\partial \boldsymbol{r}(u,v)}{\partial u}, \boldsymbol{r}_v=\frac{\partial \boldsymbol{r}(u,v)}{\partial v}$), so I guess saying "Mobius strip is not orientable" means at some $(u,v)$ of Mobius strip,
$\boldsymbol{r}_u$ or $\boldsymbol{r}_v$ is not defined, or
$|\boldsymbol{r}_u\times \boldsymbol{r}_v| = 0$
Is that right?
This definition is not quite right. A surface is orientable if a continuously-varying normal vector can be constructed at every point.
You can certainly choose a normal vector to every point in the Möbius strip, but it will not be continuously varying. What I mean is that at some point the normal vector will have to "jump" from one "side" of the surface to the other. (Or, the normal vectors will have to shrink down to $0$ in order to switch sides, but a normal vector is not allowed to be $0$.)