How can you multiply unit vectors by differentials and get this result? I'm puzzled seeing in Boas's Mathematical Method in the Physical Sciences (2nd edition, p428) that she gives
$$d\mathbf{s}=\mathbf{i}dx+\mathbf{j}dy+\mathbf{k}dz$$
$$=\mathbf{i}\left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial\theta}d\theta\right)+\mathbf{j}\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial\theta}d\theta\right)+\mathbf{k}dz.$$
But why doesn't the dual basis definition$$\boldsymbol{e}^{i}\left(\boldsymbol{e}_{j}\right)=\delta_{j}^{i}$$mean that $d\mathbf{s}=\mathbf{i}dx+\mathbf{j}dy+\mathbf{k}dz$ 
equal 3?
EDIT
She gives another similar equation on the same page:$$d\mathbf{s}=\mathbf{e}_{r}dr+\mathbf{e}_{\theta}rd\theta+\mathbf{e}_{z}dz.$$She then proceeds to square this to get $$ds^{2}=d\mathbf{s}\cdot d\mathbf{s}=dr^{2}+r^{2}d\theta^{2}+dz^{2}.$$So it's a line element. So the basis vectors squared with themselves give 1, but the differentials squared with themselves give $dr^{2}$, etc. This is probably hopelessly naive, but is it fair to say she's treating the differentials as the infinitesimal displacements of ordinary calculus rather than basis 1-forms?
BOUNTY EDIT
First, I'm confused by the author saying $$d\mathbf{s}=\mathbf{i}dx+\mathbf{j}dy+\mathbf{k}dz$$ is a vector when with all those $dx,dy,dz$ it sort of looks like a 1-form. 
Second, is @Denklo correct in saying “There seems to be a $\otimes$ (tensorial product) missing inbetween the vectors and the forms.”?
I've seen line elements written using $\otimes$ (eg, $ds\otimes ds=dt\otimes dt-dx\otimes dx-dy\otimes dy-dz\otimes dz$), but is $d\mathbf{s}=\mathbf{i}\otimes dx+\mathbf{j}\otimes dy+\mathbf{k}\otimes dz$ correct?
Third, I've found this question – If $ds$ is not a differential form, can I make sense of its intuitive notation somehow? – where @TedShifrin says “$ds$ is not a 1-form because $ds(v)=\sqrt{ds^{2}(v,v)}$ is not linear in $v$.” However, his $ds$ is not the vector $d\mathbf{s}$. 
Please don't worry about making your answers too simple – that's my level.
 A: Physicist here.

"is it fair to say she's treating the differentials as the infinitesimal displacements of ordinary calculus rather than basis 1-forms?"

Indeed, that is exactly what's going on here: this is a very common philosophy in physics. Differentials do not (usually) denote one-forms, but infinitesimal displacements. You could think of $\mathrm d\boldsymbol s$ as a small vector $\delta\boldsymbol s\to \boldsymbol 0$, whose cartesian components are $(\delta x,\delta y,\delta z)$. The formula $(\delta\boldsymbol s)^2=(\delta x)^2+(\delta y)^2+(\delta z)^2$ is just the euclidean norm of $\delta\boldsymbol s$.
With this in mind we have, for example,
$$
\delta x=x(r,\theta)-x(r+\delta r,\theta+\delta\theta)\equiv\frac{\partial x}{\partial r}\delta r+\frac{\partial x}{\partial\theta}\delta\theta+\mathcal O(\delta r,\delta\theta)^2
$$
in agreement with Boas' formula.
On the other hand, if you insist on thinking of $\mathrm d\boldsymbol s$ as a tensorial object, then you need to be more precise in your notation. If by $\boldsymbol i\,\mathrm dx$ you mean $\boldsymbol i⨼\mathrm dx$ then indeed
$$
\mathrm d\boldsymbol s=1+1+1=3
$$
But if by $\boldsymbol i\,\mathrm dx$ you mean $\boldsymbol i\otimes\mathrm dx$, then $\mathrm d\boldsymbol s$ is nothing but the identity of $V$:
$$
\mathrm d\boldsymbol s=\mathrm{id}_{V}
$$
such that $\mathrm d\boldsymbol s(\boldsymbol v)=v_x\boldsymbol i+v_y\boldsymbol j+v_z\boldsymbol z\equiv\boldsymbol v$.
