taylor expansion of scale space function
The quantity $\frac{\partial D}{\partial \textbf{x}}$ is a vector, since it is the derivative of the scalar function $D(\textbf{x})$ w.r.t. all the elements of $\textbf{x}$. In the formula it is assumed that all vectors are column vectors, so in order to compute the dot product of the derivative $\frac{\partial D}{\partial \textbf{x}}$ and the vector $\textbf{x}$, you need to transpose one of them, which gives you a matrix product row times column (= scalar). The expression $\frac{\partial D^T}{\partial \textbf{x}}$ is simply the transpose of the vector of derivatives. Maybe it would have been clearer to write it as $\left(\frac{\partial D}{\partial \textbf{x}}\right)^T$.

Similarly, for the last term you have a row vector times a matrix (second derivative of $D$ w.r.t. $\textbf{x}$) times a column vector, which again results in a scalar value.