Definitions:
A subset $P \subset \mathbb R^n$ is a polytope if it is the convex hull of finitely many points. Let $P \subset \mathbb R^n$ be a polytope. A face is a subset $F\subset P$ of the form
$$F=\arg\max\{cx : x \in P\}$$
for some $c \in \mathbb R^n$.
The dimension of a face is the dimension of its affine hull. A vertex is a zero dimensional face and an edge a one dimensional face. Two vertices $v, w$ are neighbors if their connecting line $\operatorname{conv}(\{v,w\})$ is an edge. Given a vertex $x$ define
$$N(x) = \{y \in P: \text{ $y$ is a vertex neighboring $x$}\}$$
as the set of vertices that are neighbors of $x$, and define
$$E(x) = \{y-x: y \in N(x)\}$$
as the set of edge vectors pointing from $x$ to its neighbors.
Question:
Let $P \subset \mathbb R^n$ be a polytope and let $x$ be a vertex. Let
$$E(x) = \{y-x: \text{ $y$ is a vertex neighboring $x$}\}$$ be the set
of vectors that point from $x$ to its neighboring vertices. How can we
show that for any $z \in P$ there exist coefficients $\lambda_v\ge 0$
such that $$ z = x + \sum_{v \in E(x)}\lambda_v v$$
The question can also be phrased as:
How to show that the conical hull of $P-\{x\}$,
$$K=\operatorname{cone}(P-\{x\}):=\{\sum_{i=1}^k \alpha_i (z_i-x): z_i \in P, \alpha_i\ge0, k =1,2\dots, \}$$
is generated by the edge vectors
$E(x)$ ?That is, show that
$$K=\{\sum_{y \in N(x)} \alpha_y (y-x): \alpha_i\ge0 \}.$$
See also example and images below.
I think Farkas' Lemma should lead to the answer somehow, but so far I've had no success in my proof attempts.
Example:
Consider $\mathbb R^2$ and let $P$ be the polytope that is the convex hull of the points $(0,0), (0,1), (1,0)$. If we take the vertex $x=(0,0)$ then $N(x) = \{(0,1), (1,0)\} = E(x)$ and the set of vectors that are nonnegative linear combinations of elements of $E(x)$ is $\mathbb R^2$. In particular, any $z \in P$ can be expressed as a nonnegative linear combinations of elements of $E(x)$.
Here is an image (the shaded region is the set of points $z = x + \sum_{v \in E(x)}\lambda_v v$ for some nonnegative $\lambda_v$):
Here are two more images showing the idea for different polytopes:
A polytope in $\mathbb R^2$:
A polytope in $\mathbb R^3$:
Best Answer
Farkas' Lemma is indeed the way to go, but we need the right setting. Below I give a sketch.
For simplicity, assume that we work at a vertex $x=0$ of $P$. So we want to find a minimal set of generators for the cone $\DeclareMathOperator{\cone}{cone}C:=\cone(P)=\cone (\mathcal V)$, where $\mathcal V\subseteq P$ is the set of vertices of $P$. What we want to understand is whether every such "minimal generator" $y\in\mathcal V$ is a neighbor of $x$, because if so, then the edge-directions indeed generate $C$.
So, suppose that $y\in \mathcal V$ is part of such a minimal set of generators. Then $y\not\in C':=\cone(\mathcal V\setminus \{y\})$ (here you need to use that no three vertices of $P$ are colinear). By Farkas' Lemma, we can then separate $y$ from $C'$ via a hyperplane. In particular, we can choose this hyperplane with normal vector $n$ so that
$$\def\<{\langle}\def\>{\rangle}\<n,x\>=0,\quad\<n,y\> >0\quad\text{and}\quad\<n,z\><0\text{ for all $z\in \mathcal V\setminus\{x,y\}$}.$$
It is not too hard to argue that we can choose $n$ linearly independent from $y$ (if we are working in dimension $d\ge 2$). Then
$$n':=n-y\frac{\<n,y\>}{\<y,y\>} \not=0.$$
You can check that we have $\<n',x\>=\<n',y\>=0$ and $\<n',z\><0$ for all $z\in \mathcal V\setminus\{x,y\}$ (the latter needs some thought, but is possible). In other words, the hyperplane orthogonal to $n'$ supports $P$ exacty at the two vertices $x$ and $y$, which proves that these form an edge of $P$. In still other words, $\cone(P)$ is generated by the neighbors of $x$.
Some further explanation
As requested in the comments, I elaborate on $\<n',z\><0$ for all $z\in\mathcal V\setminus\{x,y\}$. As Epiousios noted, this is the same as
$$(*)\quad \underbrace{\<n,z\>}_{<0} < \underbrace{\frac{\<n,y\>}{\<y,y\>}}_{>0} \<y,z\>,$$
which would be obviously true if $\<y,z\>>0$. However, this is not always the case.
But, we can do a trick: before we start with any of our argument, we can tranform our polytope $P$ into an more convenient polytope $P'$, for which any two neighbors $y,z$ of $x=0$ satisfy $\<y,z\>>0$ (meaning $\sphericalangle(y,z)<90^\circ$). We can do this by stretching $P$ in a certain way. Hopefully, the following image makes this clearer:
Since this is a linear transformation, this changes nothing about the actual problem. But this time $(*)$ is trivially satified.