Proof. $\blacktriangleleft$
$\boxed\implies$ If $(v_j)_1^n$ is a basis, then for each $v \in V$, there exists a unique list of scalars $(c_j)_1^n$ such that $v = \sum_1^n c_j v_j \in \bigoplus _1^n \langle v_j\rangle$, hence $V = \sum_1^n \langle v_j \rangle$. Since the expression is unique for every $v \in V$, by the defnition of direct sums, $\sum_1^n \langle v_j \rangle$ is a direct sum.
$\boxed\impliedby$ Suppose $0 = \sum_1^n c_j v_j$ for a list of scalars, then according to $c_j v_j \in \langle v_j \rangle$ for every $j$ and the definition of direct sums, $c_j = 0$ for all $j$. Hence $(v_j)_1^n$ is linearly independent. The dimension formula implies $\dim V = \sum_1^n \dim ( \langle v_j \rangle) = n$, thus $(v_j)_1^n$ shall be a basis.
$\blacktriangleright$
Appendix
Given vector space $V$, and subspaces $V_j [j =1, \dots, m]$,
$V = V_1 \oplus V_2 \oplus \cdots \oplus V_m$ is defined to be $(1)V = \sum_1^m V_j$ and $(2)$ for each $v\in V$, there are unique vectors $v_j \in V_j$ such that $v = \sum_1^m v_j$.
This is equivalent to
$V = V_1 \oplus V_2 \oplus \cdots \oplus V_m$ is defined to be $(1)V = \sum_1^m V_j$ and $(2)$ for each $k\in\{1,2,\dots,m\}$, $V_k \cap \sum_{j \neq k} V_j = \{0\}$.
Proof. $\blacktriangleleft$
$\boxed\implies$ If the expression is unique for every $v \in V$, then for $v_k \in V_k \cap \sum_{j \neq k} V_j$, there is some $u_j \in V_j [j \neq k]$ such that $v_k = \sum_{j \neq k} u_j$, hence $0 =-v_k + \sum_{j \neq k} u_j$. The uniqueness forces all $u_j$ and $v_k$ be $0$. Therefore $V_k \cap \sum_{j \neq k} = 0$ as we want.
$\boxed\impliedby$ Suppose for each $k$, $V_k \cap \sum_{j \neq k } V_j = 0$. If $0 = \sum_1^m v_j$, then $v_k = - \sum_{j \neq k} \in V_k \cap \sum_{ j \neq k } V_j = \{0\}$, so $v_k =0$ for all $k \in \{1,2,\dots,m\}$. Therefore $0$ has a unique expression $0=0+\dots + 0$. For any other $v \in V$, if $v = \sum_1^m u_j = \sum_1^m w_j$ where $u_j, w_j \in V_j$ for all $j$, then $0 = \sum_1^m (u_j - w_j)$. Since each $u_j - w_j \in V_j$ as well, the uniqueness forces $u_j = w_j$ for all $j$. Equivalently, each decomposition of $v$ is unique. $\blacktriangleright$