To answer your question at the end of the first paragraph: $u \in V$.
For the rest:
If you have a function of three variables, like
f(x, y, z) = 2x + 3y + z
you can hold two of them fixed and make a new function like this;
g(t) = f(3, t, 5).
In this case, $g(t)$ turns out to be
g(t) = 2\cdot 3 + 3t + 5 = 11 + 3t.
Is that clear? I could have set $x$ and $z$ to any values I liked; whatever values I chose, I'd get a function $g$. That function $g$ might have various properties: it might be constant. It might be linear. It might be quadrtic. It might be differentiable. And these properties might vary depending on the values you chose for $x$ and $z$ (although not in this example).
OK. You've got a function $F$ of $n$ vector variables. If you hold all but the first one fixed at values $u_2, \ldots, u_n$, you get a new function
g(t) = F(t, u_2, u_3, \ldots, u_n)
(where $t$ here is a vector rather than a real number as in the previous example).
More precisely, for any choice of $u_2 \ldots u_n$, you get a function $g$. And for any such choice, you can look at $g$ and ask "Is this a linear function of $t$?"
You could do the same thing with the second slot, and define
g(t) = F(u_1, t, u_3, \ldots, u_n)
and ask, for each possible choice of $u_1, u_3, \ldots, u_n$, whether the resulting function was linear.
And you could do the same thing for each of the other slots. If the answer is always "yes", then we say that $F$ is multilinear.
As a very simple example, the function
F(x, y) = 2xy
is a multilinear function on $\Bbb R \times \Bbb R$. You might want to check this claim carefully.
On the other hand, $F$ is not alternating, for $F(1, 1) = 2 \ne 0$. In fact, there's no alternating multilinear function on $\Bbb R \times \Bbb R$ except the constant function 0.
OP asks about "alternating"; in particular, if you have an $n$-linear function $F$ with the property that any time an argument is repeated, you get zero, why is it true that if you swap two arguments, the value negates?
To see this, I'm going to fix everything except $u_1$ and $u_2$, so I'll write
g(a, b) = F(a, b, u_3, \ldots, u_n)
and then look at the function $g$. You could do the same thing for any two arguments (the 3rd and 14th, say, or $i$th and $j$th), but making it really concrete like this helps.
Because $F$ is multilinear, we know that $g$ is bilinear, i.e.,
g(a_1 + c a_2, b) = g(a_1, b) + c g(a_2, b) \\
g(a, b_1 + c b_2) = g(a, b_1) + c g(a, b_2).
And because $F$ has the "alternating" property, we know that $g(v, v) = 0$ for any vector $v$.
Suppose that $a$ and $b$ are any vectors, and let $v = a + b$. Let's see what $g(v, v) = 0$ tells us:
0 &= g(v, v) \\
&= g(a+b, a+b) \\
&= g(a, a+b) + g(b, a + b) \\
&= g(a, a) + g(a, b) + g(b, a) + g(b, b).
Now because $g(u,u) = 0$ for any $u$, we know the first and last terms are zero, so we have
&= g(a, a) + g(a, b) + g(b, a) + g(b, b)\\
&= 0 + g(a, b) + g(b, a) + 0\\
0 &= g(a, b) + g(b, a)
from which it follows that $g(a, b) = - g(b, a)$.
So why do we define "alternating" to be "always gives zero when there's a repeated argument" instead of $g(a, b) = - g(b, a)$? Because the little proof I just gave isn't very difficult, and it's often easier to PROVE the "repeated args lead to zero" claim than the negation claim. Or maybe just because it once seemed clever to someone...I don't really know.