Hm, let me try to explain...
You can tell that the expression on the left is essentially just the limit definition of the derivative that you learned in Calc 1, except for the fact that now it's a function that takes a vector that we're taking the derivative of. The derivative is being evaluated at the point a and in the direction of the vector u. The parameter t let's you "walk" a little bit away from a in the direction of u on the function (analogous to making a step \Delta x for a single-variable function).
Then, the right hand side is just writing this same concept in a different form. It takes the derivative of the function f along our "walk" (with respect to the parameter t). Then, where do we want to find the derivative? At the location a, of course! So we have to evaluate our derivative df/dt when t=0 (when we're still at a).
In terms of actually calculating something, can't you just substitute a+tu into the function and take the derivative with respect to t?
Example:
\vec{f}(x,y) = x \hat{i} + y^2 \hat{j}
Let \vec{a} = \langle a_1, a_2 \rangle = \langle 1, 1 \rangle, \vec{u} = \langle u_1, u_2 \rangle = \langle -1, 2 \rangle
Then
\vec{f}(\vec{a} + t\vec{u}) = \vec{f}(a_1 + t u_1, a_2 + t u_2) = \vec{f}(1-t, 1+2t) = (1-t) \hat{i} + (1+2t)^2 \hat{j}
\frac{d\vec{f}(\vec{a} + t\vec{u})}{dt} \bigg|_{t=0} = -1 \hat{i} + 2 (1+2t)^2 \hat{j} \bigg|_{t=0} = -\hat{i} + 2\hat{j}
Can someone confirm this is correct? It's been a while since I've done this. The notation may not be spot on either.
