I think most of what needs to be said is in the comments, but in an attempt to set things out clearly:
If $f\colon \Omega\to \mathbb R$ is a function defined on an open set $\Omega$ of $\mathbb R^n$ (with the normal Euclidean notion of distance say) then for $a \in \Omega$ and $v \in \mathbb R^n$, the directional derivative of $f$ at $a$ in the direction $v$ is
$$
\lim_{t \to 0} \frac{f(a+t.v)-f(a)}{t}
$$
when this limit exists. It is denoted in various ways, I'll use $\partial_vf(a)$. You can check that if $r \in \mathbb R$ then $\partial_{r.v} f(a) = r.\partial_v f(a)$, that is $\partial_v f$ is compatible with scalar multiplication (so it suffices to compute directional derivatives for vectors of norm $1$ for example).
In deciding what it means for $f$ to be differentiable at $a$, there are (at least) the following possible options:
- Require only that $\partial_vf(a)$ exists for all $v \in \mathbb R^n$ -- or equivalently $\partial_v f(a)$ exists for all $v \in S^{n-1} = \{v \in \mathbb R^n: \|v\|=1\}$.
- In addition to 1., require that there is a linear map $T\colon \mathbb R^n \to \mathbb R^n$ such that $\partial_v f(a) = T(v)$.
- In addition to 1. and 2. require that $\frac{f(a+tv)-f(a)-T(t.v)}{t}\to 0$ uniformly in $v \in S^{n-1}$, that is, require $
\lim_{h \to 0}\frac{\|f(a+h)-f(a)-T(h)\|}{\|h\|}=0$.
It is easy to see that, if $f$ satisfies 3., then $\partial_vf(a) = T(v)$, and as $T$ is usually denoted $Df_a$, this becomes $\partial_v f(a) = Df_a(v)$. Using the standard basis to associate a matrix to the linear map $Df_a$, this becomes the dot product formula in the OP.
If condition 1. holds then $\partial_v f(a)$ is known as the Gateaux derivative or Gateaux differential of $f$ at $a$. Oddly, as far as I know, condition 2. doesn't seem to have a name, while condition 3., the one most widely taught, is called the Frechet derivative.
It has a number of things going for it, such as, if $f$ has a Frechet derivative at $a$ then $f$ is continuous at $a$, whereas this is false for functions satisfying conditions 1 and 2.
Examples:
i) A function which satisfies $1.$ and not $2.$ at $a=(0,0)\in\mathbb R^2$ is
$$
f_1(x,y) = \left\{ \begin{array}{cc} \frac{x^2y}{x^2+y^2} & (x,y)\neq (0,0)\\
0, & (x,y)=(0,0)\end{array}\right.
$$
Indeed $\partial_vf_1(0)= f_1(v)$.
ii) A function which satisfies 2. and not 3. is given in the comments above. If we write $1_A$ for the indicator function of $A$, that is $1_A(x)=1$ if $x\in A$ and $1_A(x)=0$ otherwise, then we get a similar example by considering $U = \{(x,y):0<y<x^2<1\}$: its indicator function $1_U$ satisfies $2.$ but not $3.$ at $a=(0,0)$. Indeed $1_U$ is of course not continuous at $(0,0)$, since $(0,0) \in \overline{U}$.
