The limit $\lim_{t\to 0} \frac{f(x_0+tv)-f(x_0)}t$ gives the definition of the derivative in the direction of the unit vector $v$ at $x=x_0\in \mathbb R^n$, that is $\frac{\partial}{\partial v} f (x_0)$.
The formula
$$\frac{\partial}{\partial v} f (x_0)=\nabla f(x_0)\cdot v$$
gives a property which is valid under the hypothesis that $f$ is differentiable at $x=x_0$, and is quite useful for calculations. (If $f$ is not differentiable at $x=x_0$, then that relation doesn't need be true, even if all directional derivatives exist.)
The idea of the proof is that being $f$ differentiable at $x_0$, then the gradient $\nabla f(x_0)$ exists and
$$\lim_{x\to x_0}\frac{|f(x)-f(x_0)-\nabla f(x_0)\cdot(x-x_0)|}{||x-x_0||}=0$$
Let's think of the point $x=x_0+tv$ (say for fixed $x_0$ and $v$). By definition of directional derivative (and substracting and adding $\nabla f(x_0)\cdot (x_0+tv-x_0$), leads to
$$\frac{\partial}{\partial v} f (x_0)=\lim_{t\to 0} \frac{f(x_0+tv)-f(x_0)}t=$$
$$=\lim_{t\to 0} \frac{f(x_0+tv)-f(x_0)-\nabla f(x_0)\cdot(x_0+tv-x_0)}{||(x_0+tv)-x_0||}\cdot \frac{|t|\,||v||}{t}+\frac{\nabla f(x_0)\cdot(x_0+tv-x_0)}{t}.$$
And because the limit of the first summand is $0$ (why?) (*) and the second one is constant the result is $$\frac{\partial}{\partial v} f (x_0)=\nabla f(x_0)\cdot v,$$
which gives the usual formula.
What might be more interesting to understand this relation is when there's no such relation. Let $f \colon \mathbb R^2 \to \mathbb R$, and
$$f(x,y)=
\begin{cases}
\tfrac{x^2y}{x^2+y^2} & (x,y)\neq (0,0) \\
0 & (x,y)=(0,0). \\
\end{cases}$$
An easy calculation using the definition shows that, if $v=(v_x,v_y)$ (let's assume $||v||=1$), the directional derivative is in each direction
$$\frac{\partial}{\partial v} f (0,0)=\frac{v_x^2 v_y}{v_x^2+v_y^2}=v_x^2 v_y$$
(in particular, both $\frac{\partial}{\partial x} f (0,0)$ and $\frac{\partial}{\partial y} f (0,0)$ are zero, that is $\nabla f(0,0)=(0,0)$.
So, if the 'dot-product formula' were valid, it should be the case that $$\frac{\partial}{\partial v} f (0,0)=(0,0)\cdot (v_x,v_y)=0,$$
which only happens in the directions of the $x$ and $y$ axes. (BTW, this also proves that $f$ is not differentiable at $(0,0)$.)
I suggest you try to imagine why the way in which directional derivatives vary as we change direction in this case (think of the $xy$ plane as the floor) are not compatible with the existence of a tangent plane (differentiability).
(*) In order to verify that
$$\lim_{t\to 0} \frac{f(x_0+tv)-f(x_0)-\nabla f(x_0)\cdot(x_0+tv-x_0)}{||(x_0+tv)-x_0||}\cdot \frac{|t|\,||v||}{t}=0,$$
first note that $\frac{|t|\,||v||}{t}$ equals plus or minus $||v||$, depending on the sign of $t$, which means is a bounded function of $t$ ($t\neq 0$). So, to prove our claim is enough to show that
$$\lim_{t\to 0} \frac{f(x_0+tv)-f(x_0)-\nabla f(x_0)\cdot(x_0+tv-x_0)}{||(x_0+tv)-x_0||}=0.$$
But this is a consequence of $f$ being differentiable. Indeed, we say that $f\colon \mathbb R^n \rightarrow \mathbb R$ is differentiable at $x_0$ if and only if
$$\lim_{x\to x_0} \frac{f(x)-f(x_0)-\nabla f(x_0)\cdot(x-x_0)}{||x-x_0||}=0.$$
Our expression just has $x_0+tv$ instead of $x$, and as the limit is for $t\to 0$, it is also true that $x_0+tv\to x_0$. The only difference is that the definition of differentiable function uses a double/triple/etc. limit (think of sequences of points of $\mathbb R^n$ converging to $x_0$ from every direction and in all sorts of simple or complicated paths), while in our limit $x$ tends to $x_0$ only along the straight line in the direction of $v$. But since $f$ is differentiable at $x_0$, the last limit is $0$, and the same is true if we restrict to the subset of $\mathbb R^n$ that is such line.
The gradient of a vector $\nabla B$ is the 2nd order (2,0) tensor.
$$
\nabla B = {\partial B_i\over\partial x_j}e_{ij}
$$
$e_{i}$ is the unit (1,0) vector basis, and $e_{ij}=e_{i}\otimes e_{j}$ is the unit matrix (2,0) basis.
Hence the product $B\cdot\nabla B$ is:
$$
B\cdot \nabla B = B_i {\partial B_i\over\partial x_j}e_{ij}
$$
For $B=B_3e_3$:
$$
B\cdot \nabla B = B_3{\partial B_3\over\partial x^{j}} e_{3j}
= B_3{\partial B_3\over\partial x^{j}} e_{j} = B_3 \nabla B_3
$$
Best Answer
Given any linear map $F\colon\Bbb R^n\longrightarrow\Bbb R$, there is one and only one vector $w\in\Bbb R^n$ such that$$(\forall v\in\Bbb R^n):F(v)=v.w.$$You just take$$w=\bigl(F(e_1),F(e_2),\ldots,F(e_n)\bigr),$$where $\{e_1,e_2,\ldots,e_n\}$ is the standard basis.
If you apply this theorem to $D_vf$, you get the existence and the unicity of the gradient.