Convex Combination - Two points Description
// Two points on space
p = [2, 3]
q = [-1, -1]

// Points between the difference of the points above
w = q - p = [-3, -4]

// Which gives us the parametric form
f(x) = (1 - t)[2, 3] + t[-3, -4]

// Sample parameter
// Which seems to overflow w = [-3, -4]
f(2) = -1[2, 3] + 2[-3, -4] = [-8, -11]

// The coefficients sum up to one
-1 + 2 = 1

As far as I can understand convex combination suffices if the sum of the coefficients are equal to 1 (l1 + l2 + ... + li = 1)
My Question
If it's a convex combination, doesn't it have to be between w = [-3, -4]?
 A: The convex combination:
$$f(t)=(1-t)(2,3)+t(-3,-4)$$
Take a point between $(2,3)$ and $(-3,-4)$ iff $t \in [0,1]$. 
If $t \notin [0,1]$ then we get a point on the same line goes through $p=(2,3)$ and $w=(-3,-4)$ but outside of the segment $\overline{pw}$.
The same idea can be generalized (for the Convex Hull) if we have $n$ points and $n$ parameters $\lambda_i$ such that $\sum \lambda_i=1$.
A: You seem to be misunderstanding the meaning of $\mathbf w$, which is not a point, but the vector pointing from $\mathbf p$ to $\mathbf q$. Let's emphasize this by putting an arrow over it: $\vec{\mathbf w}$. A convex combination of $\mathbf p$ and $\mathbf q$ is a point lying on the line segment between them, which you could parametrize in various ways, such as:
$$\mathbf x = (1-t)\mathbf p + t\mathbf q$$
or:
$$\mathbf x = \mathbf p + t\vec{\mathbf w}$$
Note that by substituting $\vec{\mathbf w} = \mathbf q - \mathbf p$ in the second formula, we get the first, so they are equivalent. Observe also that when $t=0$ or $1$, $\mathbf x = \mathbf p$ or $\mathbf q$ accordingly. Finally, to make sure that it is a convex combination (and not just a linear combination), we add the restriction $0 \le t \le 1$.
Your question ("Doesn't it have to be between $\mathbf w = (-3,-4)$?") is also a bit strange: the concept of "between" requires two objects, not just one.
