Constructing a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces

It is not even possible for $m = n-1$.

Let $E = \{ e_1,\dots,e_n \}$ be the standard basis of $\mathbb{R}^n$. Let $V$ be the the subspace generated by $\{ e_1,\dots,e_{n-1} \}$ (i.e. $V = \mathbb{R}^{n-1} \times \{ 0\}$) and $V_j = V$ for $j =1,\dots,n-1$. Let $V_n$ be the subspace generated by $\{ e_1,\dots,e_{n-2},e_n \}$. Finally let $U = E$ and $W = \{ e_1,\dots,-e_{n-1},-e_n \}$. These are two bases of $\mathbb{R}^n$ which have the same orientation.

Assume there exist paths $\gamma_j$ as desired. Then $\gamma_n$ is a path in $V_n \subset \mathbb{R}^n$ such that $\gamma_n(0) = e_n$, $\gamma_n(1) = -e_n$. If $p_n : \mathbb{R}^n \to \mathbb{R}$ denotes the projection $p_n(x_1,\dots,x_n) = x_n$, we see that $p_n\gamma_n(0) = 1$ and $p_n\gamma_n(1) = -1$. Since $p_n\gamma_n$ is continuous, there exists $t \in I$ such that $p_n\gamma_n(t) = 0$ which means $\gamma_n(t) \in V$. Hence $\gamma(t) \in V$. This shows that $\gamma(t)$ cannot be a basis of $\mathbb{R}^n$ which is a contradiction.

In general it is not true. Let $U =\{u_1,\dots,u_n \}$ be a basis of $\mathbb{R}^n$ and $V_j$ be the one-dimensional subspace generated by $u_j$. But now for any collection $\epsilon_j \in \{ 0,1 \}$ the $w_j = (-1)^{\epsilon_j}u_j$ form again a basis of $\mathbb{R}^n$, andf if $\epsilon_j = 1$, then any continuous path in $V_j$ connecting $u_j$ and $w_j$ goes through $0$ so that $\gamma(t)$ cannot be a basis for all $t$.

Edited:

As a first step beyond $m=1$ it seems reasonable to study the case $n= 3, m=2$.

Morerover, if the assertion is true for $m=2$, then it also true for all $m =2,\dots,n-1$ (because for each $m > 2$ there exist two-dimensional subspaces $V'_j \subset V_j$ such that $u_j,w_j \in V'_j$).