The matrix of the operator $f$ with respect to the basis $B_c$ (is this what your notation $[F]_C$ refers to?) tells you the values of $f$ on this basis. For example, the first column tells you that
$f(e_1) = 1\cdot e_1 + 0\cdot e_2 + (-1)\cdot e_3 = e_1-e_3.$
Then if we want to compute the value of $f(x,y,z)$ for an arbitrary vector $(x,y,z)\in\mathbb{R}^3$, it would be useful to decompose this vector in terms of $B_c$ and then exploit the linearity of $f$.
Now, the notation $e_i$ usually refers to the standard basis vector of $\mathbb{R}^n$ whose entries are all zero except for the $i$th entry, which is 1. In the case above, we have $e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1).$
An arbitrary vector $(x,y,z)$ can be written as a linear combination of this particular basis in a simple way:
$(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1)$
$= xe_1 + ye_2 + ze_3.$
So if we want to calculate $f(x,y,z)$, just apply $f$ to both sides:
$f(x,y,z) = f( xe_1 + ye_2 + ze_3).$
Now using the fact that $f$ is a linear operator, we can expand the right-hand side:
$f(x,y,z) = xf(e_1) + yf(e_2) + zf(e_3).$
From here, you can read off the values $f(e_i)$ from the matrix.
To answer your question about why we don't apply $f$ to the numbers $x,y,z$: the linear operator $f$ is a function from $\mathbb{R}^3$ to $\mathbb{R}^3$. So it only makes sense to evaluate $f(v)$ where $v$ is an element of $\mathbb{R}^3$, i.e. a triple of real numbers, rather than a single real number. However, the fact that $f$ is a linear operator means that it is compatible with multiplication by real numbers, in the sense that
$f(cv) = cf(v)$ whenever $c$ is a real number and $v$ is an element of $\mathbb{R}^3$. Linearity also means that $f(v+w) = f(v) + f(w)$ for all $v,w\in\mathbb{R}^3$. Combining these two facts is how the right-hand side of the above equation was expanded.