For a counterexample:
Let $V=\mathbb R^2$ and $X=\{(x,0):x\in\mathbb R\}$ and $Y=\{(0,y):y\in\mathbb R\}$.
Then $X$ and $Y$ are subspaces of $V$.
Let $I$ be the identity operator on $V$.
You can see that $I(X\oplus Y)\neq I(X)\cup I(Y)$.
To make your statement true you can have:
Given 2 subspaces of a vector space $V$ and a linear operator on $V$, the image of the direct sum of the subspaces is equal to the sum of the images under the operator.