Decomposition of vector space by intersections

If a vector space has two deompositions $$V = V_1 \oplus V_2 = W_1 \oplus W_2$$, then is it true that $$V = \oplus_{i,j=1,2} V_i \cap W_j$$

Of course, I guess so. But the following example is not so. Probably I have misunderstand but I cannot found it. Please let me know where is wrong.

Let $$V$$ be a three dimensional vector space and denote its basis by $$e_1, e_2,e_3$$.

Take another basis $$f_1,f_2,f_3$$ constructed from $$e_1, e_2,e_3$$ as follows. \begin{align*} f_1 &= e_1 +e_2,\\ f_2 &= e_3,\\ f_3 &= e_2 +e_3. \end{align*}

This transformation matrix has non-zero determinant.

Consider two decompositions of V, namely,

$$V= (e_1,e_2) \oplus (e_3)=: V_1 \oplus V_2,$$

$$V= (f_1,f_2) \oplus (f_3)=: W_1 \oplus W_2.$$

From the following calculations

$$V_1 \cap W_1 = (f_1),$$

$$V_1 \cap W_2 = 0,$$

$$V_2 \cap W_1 = (e_3),$$

$$V_2 \cap W_2 = 0,$$

I cannot deduce $$V = \oplus_{i,j=1,2} V_i \cap W_j$$

Indeed your desired formula does not hold. Consider for example the two bases $$\{(1,0),(0,1)\}$$ and $$\{(1,0),(1,1)\}$$ of $$K^2$$, where $$K$$ denotes some field. Then \begin{align*} K\cdot (1,0) \cap K\cdot (1,0) &= K \cdot (1,0)\\ K\cdot (1,0) \cap K\cdot (1,1) &= \{0\}\\ K \cdot (0,1) \cap K \cdot (1,0) &= \{0\}\\ K \cdot (0,1) \cap K \cdot (1,1) &= \{0\} \end{align*} cannot possibly sum up to form $$K^2$$...
This proves in particular, that any vector space of dimension $$\geq 2$$ admits two such decompositions, which cannot be refined to make your formula hold.