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.
If you define $\nabla_x f(x_0)=\lim_{h \to 0^+} \frac{f(x_0+hx)-f(x_0)}{h}$, then you have the identity $\nabla_x f(x_0)=\| x \| \nabla_{x/\| x \|} f(x_0)$. (I will remark that this notation clashes with notation elsewhere in math, but I will stick with it here.) That is, the derivative "along $x$" is the directional derivative multiplied by the norm of $x$. In effect instead of just moving in a direction and measuring the change in $f$ relative to the distance you traveled in that direction, you are moving in a direction at a particular rate in time and measuring the change in $f$ relative to that change in time. The speed is the conversion factor between these measurements.
This definition of $\nabla_x$ doesn't depend on there being such a thing as the norm of $x$, whereas the directional derivative does. But for your purposes you can ignore this remark for now.
I said this in the first paragraph, but just to directly address your third question, let me add one more thing. The directional derivative does not really have a notion of time, it is really a change in $f$ with respect to distance traveled in the specified direction. Your generalized notion $\nabla_x$ effectively involves time after you identify $\| x \|$ as a speed and $h$ as a time, so that $hx$ is a displacement and $h \| x \|$ is a length.
Best Answer
Let $f: D\subseteq \Bbb R^n \to \Bbb R$ be differentiable. Then
$$\begin{align}\require{cancel}\nabla_{\vec v} f(\vec x_0) &= \lim_{h\to 0} \frac{f(\vec x_0 + h\vec v)-f(\vec x_0)}{h} \\ &= \lim_{h\to 0} \frac{\left(\color{red}{\cancel {\color{black}{f(\vec x_0)}}}+\nabla f(\vec x_0)\cdot (h\vec v) + o(h)\right) - \color{red}{\cancel {\color{black}{f(\vec x_0)}}}}{h} \\ &= \lim_{h\to 0} \frac{\color{red}{\cancel {\color{black}{h}}}\left[\nabla f(\vec x_0)\cdot (\vec v)\right]}{\color{red}{\cancel {\color{black}{h}}}} + \cancelto{0}{\lim_{h\to 0}\frac{o(h)}{h}} \\ &= \nabla f(\vec x_0) \cdot \vec v\end{align}$$
You don't have to use a unit vector to calculate the directional derivative, but the dd will only correspond to the geometric idea of slope if you use a unit vector $\vec v$.
Edit: I assume that you are familiar with Taylor's theorem. Recall that the first order Taylor expansion of a function $g: \Bbb R\to \Bbb R$ around $a$ is
$$g(a+h) = g(a) + g'(a)h + o(h)$$
Here $o(h)$ is a stand-in for the remainder function $g(a+h)-g(a)-g'(a)h$. This notation (called little oh notation) tells us that the remainder has the property $$\lim_{h\to 0}\frac{g(a+h)-g(a)-g'(a)h}{h} = 0.$$
For functions of a vector variable, there's a similar Taylor expansion:
$$f(\vec a + \vec h) = f(\vec a) + \nabla f(\vec a)\cdot \vec h + o(\|\vec h\|)$$
So what I'm doing above is replacing $f(\vec x_0 + h\vec v)$ with its first order Taylor expansion. Then two terms cancel, one tends to zero, and we're left with the identity you're looking for.