Let V be a finite dimensional vector space over an arbitrary field. Let $U_1,U_2$ be subspaces of $V$ and $B_1,B_2$ be any bases for $U_1$ and $U_2$. Then V is the direct sum of $U_1$ and $U_2$ if and only if $B_1\cup B_2$ is a basis for $V$ and $B_1 \cap B_2$ $=$ $\varnothing$
I will prove the converse. So I will let $\{$ $u_1,u_2,u_3....,u_n$ $\}$ and $\{$ $w_1,w_2,....w_m$ $\}$ be bases for $U_1$ and $U_2$ respectively. I will further suppose that $B_1\cup B_2$ is a basis for $V$ and $B_1 \cap B_2$ $=$ $\varnothing$. Note that as $B_1\cup B_2$ is a basis, then $v=u_1+u_2$ for some $u_1\in U_1$ and $u_2\in U_2$ ( since span($B_i)=U_i$). Thus, it suffices to show that any vector $v\in V$ can be expressed uniquely as $u_1+u_2$ where $u_i\in U_i$ for $i=1,2$. Note that as $B_1\cap B_2 = \varnothing$ then the coefficients when we express $v\in V$ as elements of $B_1\cup B_2$ must be unique and so each v can be expressed uniquely.
Is the proof above correct? My issue with the above proof is that if $B_1 \cap B_2\neq \varnothing$ then $B_1\cup B_2$ cannot even be a basis, so isn't it enough to just assume that $B_1\cup B_2$ is a basis for $V$?