Can the directional derivative be thought of as a rate of change of function with respect to some arbitrary paramter on which all its inputs depend? The formal definition of a directional derivative is 
$$\nabla_{\vec{v}}f(\vec{a}) = \lim_{h \to 0}\frac{f(\vec{a} + h\vec{v}) - f(\vec{a})}{h |\vec{v}|}$$
Let's take a function with constant x and y slopes for an example, (but this reasoning works well even if the slopes are just a function of x and y, because at a given point they will have a constant value)
$$f(x, y) = x + y$$
So its partial derivatives with x and y are $1$ and $1$ respectively.
Let's take an variable $t$ such that the inputs of our function depend on it in any arbitrary manner, but to keep things simple I've chosen a simple example
$$x (t) = t, y(t) = t$$
So my understanding is that the "directional" derivative is the answer to the natural question that arises that what is the rate of change of $f$ as we change $t$
And we can get the derivative by simple substitution and derivating the function w.r.t. t, we get $f_t = 2$
Now here's where a contradiction is arising, because the formally defined directional derivative in the direction along the plane $x = y$ would be $\sqrt{2}$.
My resolution to this contradiction is that if $x (t) = t, y(t) = t$ then indeed $\frac {df} {dt} = 2$.
But if there were a $v$ that was just $t$ scaled to satisfy the condition $(v(t))^2 = (x(t))^2 + (y(t))^2$ at the given point, i.e. for this example 
$$x(v) = \frac {v} {\sqrt{2}}, y(v) = \frac {v} {\sqrt {2}}$$
Then we will get 
$$\frac {df} {dv} = \sqrt {2}$$
Which is the same as the "directional derivative" of the function $f$ in direction along the plane $x = y$
Now why I like this mental image of what's going on is because it essentially lets me transform the given function $f$ to become a single parameter function provided that all its original inputs depend on that single parameter. Then the result of the derivative becomes useful as rate of change of that function with respect to that parameter.
But is this way of thinking correct? Is there something about directional derivatives that cannot fall under this analogy? Is the reasoning about scaling to the vector-like condition of $(v(t))^2 = (x(t))^2 + (y(t))^2$ correct? (it seems a little ad-hoc)
 A: The direction vector $v$ is a unit vector:
$$
v = (1/\sqrt{2}) \, (1,1)^t
$$
and as $f$ has a total derivative we get the directional derivative
$$
\partial f / \partial v =
\DeclareMathOperator{grad}{grad}
\grad f \cdot v = (1,1)^t \cdot (1/\sqrt{2}) \, (1,1)^t = (1 / \sqrt{2}) \, 2 = \sqrt{2}
$$
Checking the change of $f$ for the unit step in $v$-direction:
$$
\frac{\Delta f}{h}
= \frac{f(1/\sqrt{2}, 1/\sqrt{2}) - f(0,0)}{1} = 2/\sqrt{2} = \sqrt{2}
$$
so I do not see the contradiction you mention. 
You did calculate
$$
\frac{d}{dt} f(t, t) = \frac{d}{dt} 2t = 2
$$
which is not the same as $du = v dt$ and
$$
df = \grad f \cdot du 
= \grad f \cdot (v \, dt) 
= (\grad f \cdot v) \, dt = (\partial f / \partial v) dt
$$
This is mainly because by increasing $t$ by $1$ you do not travel a unit step along the line $\alpha (1,1)^t$ but rather a distance of $\sqrt{2}$. So your rate of change for $f$ is increased by that factor.
A: You can choose a path but in the end (limit $t \to 0$), there will be a tangent line (if the derivative exists) in a certain direction. That direction is defined up to scaling and if you don't compensate for this in the definition of the directional derivative, it is not meaningful to compare values.
You take $(x,y) = (t,t)$ and get a value of $2$ after substitution; but the same path can be parametrized as $(x,y) = (2t,2t)$ and then you would get a value of $4$, although you've calculated a rate of change in the same direction. To compare values in a meaningful way, you have to use vectors of the same length. Since the normal partial derivatives implicitly use unit vectors (when viewed as a specific case of the directional derivative) and because this is the easiest / most straightforward choice, we always take unit vectors and / or normalize to compensate for it.
Or to continue on your example: the partial derivative for your function w.r.t. $x$ is $1$, but if you would parametrize a path $(x,y)=(2t,0)$, you would find the value $2$ after substitution and differentiation... If you divide by the norm, then no problem (or conflict with the definition of the directional derivative) arises.
A: If we parameterize $x$ and $y$ as both equal to $t$, we are not guaranteed that the derivative with respect to $t$ equals the derivative with respect to $x$ or $y$.
Using the rule for total derivative:
$$
\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{d x}{dt}+
\frac{\partial f}{\partial y}\frac{d y}{dt}.
$$
A: You shall choose $x=t\cos\alpha,\ y=t\sin\alpha$, with $\alpha$ being a constant that individuates the direction, and derivate vs. $t$.
That is the same as $h\vec v$ in the introduction, but $\vec v$ being a unitary vector, to keep congruency with $d/dx$.
