Understanding for alternating functions $f$ is alternating if $f(v_1,...,v_n)=0$ whenever $v_i=v_j$ for some $i\neq j$.
Can you explain this definition more clearly? 
 A: The definition you give,

$f$ is alternating if $f(v_1,...,v_n)=0$ whenever $v_i=v_j$ for some $i\ne j$,

is specifically used for multilinear functions, that is, functions which are linear in every argument. For such function it is equivalent to the following, more obvious definition:

$f$ is alternating if it is changes sign whenever two arguments are exchanged.

To see how this works, let's look at a function with just two arguments, $f(u,v)$.
It is immediately obvious that for the second definition, for $u=v$ we get $f(\color{red}{u},\color{blue}{u})=-f(\color{blue}{u},\color{red}{u})$ (note that the colours are there to help understanding what happens, but the different-coloured variables still denote the same value, that is, $\color{red}{u}=\color{blue}{u}=u$). But $f(u,u)=-f(u,u)$ means $f(u,u)=0$.
For the other direction, we need to explicitly use linearity in both arguments. Due to linearity, we have
$$f(u+v,u+v) = f(u,u) + f(u,v) + f(v,u) + f(v,v)$$
But according to the definition you cited, $f(u+v,u+v)$, $f(u,u)$ and $f(v,v)$ all vanish, so that equation reduces to
$$0 = f(u,v) + f(v,u),$$
that is,
$$f(u,v) = -f(v,u)$$
A: Other users have provided examples of alternating functions and I see you still have a question about what does it mean for a function to be alternating. A function $f(v_1,...,v_n) \in \mathbb{R}$ is said to be alternating if $f(v_{\sigma(1)},...,v_{\sigma(n)}) = \textbf{sgn}(\sigma) \cdot f(v_1,...,v_n)$ where $\sigma \in (S_n, \circ)$, $(S_n, \circ)$ is the symmetric group of $A \equiv \lbrace 1, 2, 3, \cdots \cdots, n \rbrace$ and $\textbf{sgn}(\sigma) = \pm 1$. I see your definition as more of a corollary of the above definition (I show this below) and this is probably why it was difficult to come up with non-trivial examples.
What the above definition means is that, no matter how you shuffle around the input variables, you always get the original functions value back, just with a $\pm 1$ in front. If $\sigma \in S_n$ then $\sigma$ is a permutation of $\{1,..,n\}$ and so it shuffles the input variables around. Before moving on, let us define $\sigma (f) = f(v_{\sigma(1)},...,v_{\sigma(n)})$. Since $f$ is alternating, then $\sigma(f) = \pm  f$. 
Recall that every permutation is a product of simple transpositions (i.e $\sigma = \tau_1 \cdots \tau_k$), and so to undo the shuffling, we take $\sigma^{-1} = \tau_k^{-1} \cdots \tau_1^{-1}$. Hence, $\textbf{sgn}(\sigma) = (-1)^k$.  Thus if $v_i = v_j$ and $j \not = i$ then $\sigma = (i,j)$ and $\sigma(f) = f = - f$ which implies $f = 0$. I hope this helps.
