Dot products in spherical polars?

The scalar product is bilinear. It means that if you take a vector $\overrightarrow{a} = A_1 \overrightarrow{a_1} + A_2 \overrightarrow{a_2}$ and $\overrightarrow{b} = B_1 \overrightarrow{b_1} + B_2 \overrightarrow{b_2}$, the scalar product will write: \begin{align} \overrightarrow{a} \cdot \overrightarrow{b} &= (A_1 \overrightarrow{a_1} + A_2 \overrightarrow{a_2}) \cdot (B_1 \overrightarrow{b_1} + B_2 \overrightarrow{b_2}) \\ &= A_1 B_1 \overrightarrow{a_1} \cdot \overrightarrow{b_1} + A_1 B_2 \overrightarrow{a_1} \cdot \overrightarrow{b_2} + A_2 B_1 \overrightarrow{a_2} \cdot \overrightarrow{b_1} + A_2 B_2 \overrightarrow{a_2} \cdot \overrightarrow{b_2}. \end{align}

Now what happens in your case is that $\overrightarrow{a}$ and $\overrightarrow{b}$ are decomposed over the same orthonormal basis (vectors have a unit length and are "mutually perpendicular"). Two vectors being "perpendicular" means nothing more than the fact that their dot product is $0$. So if you develop your expression to let appear all the terms just as I did in my example, you will find the usual $a_r b_r + a_{\theta} b_{\theta} + a_{\phi} b_{\phi}$ (just like in Cartesian coordinates), plus all the terms proportional to $\overrightarrow{e_r} \cdot \overrightarrow{e_{\theta}}$, $\overrightarrow{e_r} \cdot \overrightarrow{e_{\phi}}$ and $\overrightarrow{e_{\theta}} \cdot \overrightarrow{e_{\phi}}$, that will be equal to $0$ because $\overrightarrow{e_r}, ~ \overrightarrow{e_{\theta}}$ and $\overrightarrow{e_{\phi}}$ are mutually perpendicular.

The result stems from the fact that the spherical coordinates are orthogonal (i.e., mutually perpendicular), which makes the unit vectors orthonormal, so we should have that $$\mathrm{e}_i\cdot\mathrm{e}_j=\begin{cases}1&\text{for }i=j \\ 0&\text{otherwise}\end{cases}\tag{1}$$ where $\mathrm e_i$ is the unit vector.

Another way of seeing this is through another common definition of the dot product as the magnitude of the two vectors times the angle between them: $$a\cdot b=\Vert a\Vert \,\Vert b\Vert \cos\delta $$ Perpendicular (orthogonal) axes have $\delta=90^\circ$ while aligned axes have $\delta=0^\circ$, which means that, $$ \mathrm{e}_i\cdot\mathrm{e}_j=\begin{cases}\cos0^\circ\equiv1&\text{for }i=j \\ \cos90^\circ\equiv0&\text{otherwise}\end{cases} $$ which is the same as (1) above.