Let $V$ and $W$ be two finite-dimensional real vector spaces. I guess that there is only one way to interpret the title (if my guess is wrong, then please let me know), but here is what I mean:
Suppose we have chosen an orientation for $V$ and an orientation for $W$. We can choose a positive basis $v_1,\ldots,v_m$ of $V$ and a positive basis $w_1,\ldots,w_n$ of $W$. If the orientation of the basis $$e_1,\ldots,e_{m+n}:=(v_1,0),\ldots,(v_m,0),(0,w_1),\ldots,(0,w_n)$$turns out to be independent of the chosen bases, then this determines an orientation of $V\oplus W$ in dependence of the chosen orientations of $V$ and $W$.
I think that the answer is YES. Here is my attempt of a proof: We want to show that $$e_1,\ldots,e_{m+n}:=(v_1,0),\ldots,(v_m,0),(0,w_1),\ldots,(0,w_n)$$ and $$\tilde e_1,\ldots,\tilde e_{m+n}:=(\tilde v_1,0),\ldots,(\tilde v_m,0),(0,\tilde w_1),\ldots,(0,\tilde w_n)$$ have the same orientations if
- $v_1,\ldots,v_m$ and $\tilde v_1,\ldots,\tilde v_m$ have the same orientation and
- $w_1,\ldots,w_n$ and $\tilde w_1,\ldots,\tilde w_n$ have the same orientation.
Set $M_{ij}=e^i\tilde e_j$, $A_{ij}=v^i\tilde v_j$ and $B_{ij}=w^i\tilde w_j$, then I assume that \begin{equation} M=\begin{pmatrix}A&0\\ 0&B\end{pmatrix} \end{equation} and hence $\det M=\det A\det B>0$, which shows that $e$ and $\tilde e$ have the same orientation.