Suppose $X$ is a finite-dimensional linear space, $U$ and $V$ two subspaces of $X$ such that $X$ is the sum of $U$ and $V$, i.e, $X=U+V$. Denote by $W$ the intersection of $U$ and $V$, i.e., $W=U\cap V$. Define $U_0=U/W$ and $V_0=V/W$. Then $U_0 \cap V_0=\{0\}$
I do not quite understand the last sentence.
In my mind: $U_0=\{u+W|u\in U\}$ and $V_0=\{v+W|v\in V\}$ Therefore, $U_0 \cap V_0$ can even equal to $X$ if $U=V=X$
Your notation for the quotient space is a little vague. Perhaps this description will help. The quotient space is the space obtained by shrinking the entire subspace $W$ to a single point $0$. The resulting space is $X/W$, and it is a distinct space from $X$ (it is not a subspace). So when we talk about $U_0 \cap V_0$, $U_0 = U/W$ and $V_0 = V/W$ are subspaces of $X/W$, not of $X$. Therefore we certainly could not have $U_0 \cap V_0 = X$, since $X$ is an entirely different space. The most we can ask for is $U_0 \cap V_0 = X/W$.
If you have $U = V = X$, then $W = U \cap V = X$ and $U/W = V/W = X/W = X/X = \{0\}$. So it is indeed true that $U_0 \cap V_0 = X/W$ (not $= X$), but also that $U_0 \cap V_0 = \{0\}$.