A vector function can act on multiple variables by having each component function act on multiple variables. If f defines the electric field throughout a transformer, It takes three input variables, specifying a point inside the volume of the transformer, and produces three output variables, the strength of the electric field along each of the three axes at the specified location.
When the number of input variables equals the number of output variables the vector function is sometimes called a vector field. This is derived from terms in physics: electric field, magnetic field, gravitational field, etc. Vector fields usually describe the "push" experienced by an object for each point in space. The push is moving you to some other point in space, magnitude and direction, so there are 3 input variables and 3 output variables.
Let's consider the derivative of a vector function. As a charged particle moves in the x direction inside our hypothetical transformer, how is the force on that electron changing, as measured along the x axis? This is the first partial of the first component function. If the electron moves parallel to the y axis, the change in force in the x direction is the second partial of the first component function. We have three coordinates and three component functions, hence 9 entries, arranged in a 3 by 3 matrix.
In general, a function f with m input variables and n output variables produces an n by m matrix, called the jacobian of f. the ith row of the matrix lists the m partial derivatives of the ith component function. There are n rows, one for each component function in f.
The jacobian is the derivative of f, but that's not by definition. We need to prove it. First note that the vector function f is continuous iff its component functions are continuous. We won't prove it in detail, but it's pretty straightforward. If everything near p maps to a small box centered at f(p), each component is restricted to the side of that small box. Conversely, if the component functions can all be restricted to tiny segmens centered at p, intersect their neighborhoods and the entire function is restricted to a tiny box around p.
The vector function f is differentiable at p if there is a linear approximation to f, in some neighborhood about p, and the error term divided by the distance goes to zero as we approach p. The linear approximation is an n by m matrix of values, since we must define the change in each output variable in response to each input variable.
Let f1 be the first component function. The linear approximation of f implies a linear approximation of f1. This is the first row of the matrix. If this approximation is not sufficiently close to f1, then the entire approximation isn't going to be close to f. (The first coordinate is always too far off the mark.) Hence the first row of the matrix must be a good linear approximation of f1, in other words, the derivative of f1. The same holds for all the other component functions. If f is differentiable its derivative has to be the individual derivatives, i.e. the gradients, of the component functions.
Conversely, if the components all have good linear approximations, then so does f. The composite error term is no more than the sum of the individual error terms, which drop to zero. Thus f is differentiable iff its components are differentiable, and the jacobian has to be the derivative of f.
If we start at p and travel along a certain vector v, how is f changing? The change in f is simply the changes in the component functions, pasted together, and the directional derivative can be computed per component. Putting this all together, the change in all the coordinates of f is given by the jacobian matrix on the left, times the direction vector v on the right (a column vector), using matrix multiplication. The resulting vector gives the change in each of the n coordinates of f as you move in the direction of v.
Partials are special directional derivatives, as you move parallel to the coordinate axes. They are represented by the columns of the jacobian.