Think about differentiation in one variable for a second. The derivative of the one-dimensional vector function (from $\mathbb R$ to $\mathbb R$) gives you the slope at each point. Another way to look at it is that at each point $x \in \mathbb R$ you associate to it a linear transform, f'(x) : \mathbb R \to \mathbb R which for $y$ gives you the number f'(x)y (here this is just standard number multiplication). The transform way is the right way to see things in $\mathbb R^n \to \mathbb R^m$ and furthermore from $V \to W$ where $V$ and $W$ are arbitrary normed vector spaces.
Now when you think of the gradient, say $f : \mathbb R^n \to \mathbb R$, you have a differentiation operator $\nabla : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R)$ (the latter means all linear applications from $\mathbb R^n$ to $\mathbb R$) which maps every $\mathbb x \in \mathbb R^n$ to $\nabla f(x)$, but instead of seeing $\nabla f(x)$ as a vector, think of it as a linear transform : $\nabla f(x) : \mathbb R^n \to \mathbb R$ and $\nabla f(x)(y) = \nabla f(x) \cdot y$, where on the left you apply the function $\nabla f(x)$ on $y$ and on the right this is the scalar product of both vectors.
When $f : \mathbb R^n \to \mathbb R^m$, exactly the same thing happens : you have a differentation operator $D : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R^m)$ and $x \in \mathbb R^n$ gets mapped to $Df(x)$ (here the old $\nabla$ became the more general $D$), the matrix with the partial derivatives in it. When you compute $Df(x)(y)$, this is just matrix multiplication : $Df(x)$ is the $m \times n$ matrix and $y$ is a $n \times 1$ column vector.
Note that it isn't true that $Df(\vec{\xi})(\vec x) \cdot \vec x = 0$ in general : think of the case where $f$ is the identity function, thus $Df(\vec{\xi})$ is the identity matrix for every $\vec{x}$, hence $Df(\vec{\xi})(\vec x) \cdot \vec x = \vec x \neq 0$ as long as $\vec x \neq 0$. Also note that the $\circ$ you had put up there is not the scalar product but rather matrix multiplication when $f : \mathbb R^n \to \mathbb R^m$.
Hope that helps,