For $ V = A_1 \oplus A_2 \oplus ... \oplus A_k $ , where $ A_1, A_2,...,A_k$ are subspaces of $ V $.
Explain why for $ \sum_{i=1}^k a_i = 0 $ with each element $a_i \ \epsilon \ A_i $ we can deduce each $a_i = 0 $.
I am assuming that you can start with the statement,
$ A_1 \bigcap A_2 \bigcap ... \bigcap A_k = \{0\}$
but don't know how to take this from here.
On
It depends on how you define for a set of subspaces to give a direct sum.
I usually define $A_1,\dots,A_k$ to be independent subspaces of $V$ if (and only if) from $a_i\in A_i$ and $a_1+\dots+a_k$, it follows that $a_1=a_2=\dots=a_k=0$.
In this case we can write $A_1+\dots+A_k=A_1\oplus\dots\oplus A_k$.
An equivalent statement is that, for $i=1,2,\dots,k$, $$ A_i\cap \biggl(\, \sum_{\substack{1\le j\le k\\j\ne i}}A_j\biggr)=\{0\} $$ It is not sufficient to assume $A_i\cap A_j=\{0\}$, for $i\ne j$, nor $A_1\cap A_2\cap\dots\cap A_k=\{0\}$ (which is weaker).
For instance, if $V=\mathbb{R}^2$ and we set $A_1=\langle e_1\rangle$, $A_2=\langle e_2\rangle$, $A_3=\langle e_1+e_2\rangle$, we have $A_i\cap A_j=\{0\}$ for $i\ne j$, but the three subspaces aren't independent, since $e_1+e_2-(e_1+e_2)=0$.
For finite dimensional spaces, independence is equivalent to $$ \dim(A_1+\dots+A_k)=\sum_{1\le i\le k}\dim A_i $$
You should assume $A_i \cap \oplus_{j\neq i} A_j =\{0\}$ (or more precisely $A_i \cap \Sigma_{j\neq i} A_j =\{0\}$) for any $i$ which is in fact equivalent to the statement afterwards.