I'm struggling to see how these two definitions of geometric independence are related.
In Elements of Algebraic Topology by J. Munkres the following definition is given:
Given a set $\{a_0,a_1,\ldots,a_k\}$ of points in $\mathbb{R}^n$ is said to be geometrically independent if for any (real) scalars $t_i$, the equations \begin{equation} \sum_{i=0}^kt_i =0, \text{ and }\sum_{i=0}^k t_ia_i =\boldsymbol{0} \end{equation} imply that $t_0 = t_1 = \ldots = t_k =0$.
And in Basic Concepts of Algebraic Topology, by F. H. Croom, geometric independence is defined as
A set $A=\{a_0,a_1,\ldots,a_k\}$ of $k+1$ points in $\mathbb{R}^n$ is geometrically independent means that no hyperplane of dimension $k-1$ contains all the points.
Croom defines the hyperplane as the set $H = \{v+a\mid a\in A\}$, where $A$ is a subspace of a vector space $V$ and $v\in V$.
Any help is gladly appreciated.
As pointed out in the link, both are equivalent to the vectors $v_1=a_1-a_0, \ldots, v_k=a_k-a_0$ being linearly independent.
For the Croom definition this is more or less clear (a hyperplane through the $a$s can be translated to a subspace through the $v$s and vice versa).
For the Munkres definition: Supposing $\sum c_i v_i=0$ gives $\sum t_i a_i=0$ with $t_0=-\sum c_i$ and $t_i = c_i$ for other $i$. If $a$s are independent, then all $t_i=0$. Thus $c_i$s are, too. So $v$s are independent. Conversly, having $\sum t_i a_i=0$ with $\sum t_i=0$ we can take $c_i=t_i$ for $i>0$ and get $\sum c_i v_i=0$. So if $v$s are independent, then $c$s are zero, and hence so are $t$s.