Here is a part of an exercise from Guillemin-Pollack:
Which of the following linear subspaces intersect transversally?
(d) $R^k\times \{0\}$ and $\{0\} \times R^l$ in $R^n$. (Depends on $k,l,n$.)
(e) $R^k\times \{0\}$ and $ R^l\times \{0\}$ in $R^n$. (Depends on $k,l,n$.)
(f) $V\times \{0\}$ and the diagonal $\Delta$ in $V\times V$.
What's the conceptual difference between (d) and (e)? It seems to me that the answer in both cases is the same, namely the intersection is transversal iff $k+l\ge n$. Just because two subspaces intersect transversally iff the sum of their dimensions is greater than or equal to the dimension of the ambient space. Is that correct?
For (f), suppose $\dim V=n$. Then $\dim (V\times \{0\})=n; \dim \Delta=n; \dim (V\times V)=2n$. Since $\dim (V\times \{0\})+\dim \Delta=\dim (V\times V)$, the intersection is transversal. Is that correct?
The difference is crucial, when you write $$\mathbb{R}^k\times \{0\}\qquad\text{ in }\mathbb{R}^n$$ you really mean $$\{(x_1,\ldots,x_k,0,\ldots, 0)\in \mathbb{R}^n\}$$ whereas $$\{0\}\times \mathbb{R}^k\qquad\text{ in }\mathbb{R}^n$$ means $$\{(0,\ldots, 0, x_{n-k+1},\ldots,x_n)\in \mathbb{R}^n\}$$ Therefore in case (e), there is always one subspace that contains the other. It is therefore much more difficult to be in transverse position, the bigger one has to be the whole of $\mathbb{R}^n$.