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)

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.
