The shortest walk must have the form
u
\
t<->w
/
v
where the edges in each arrow are disjoint. To see this, suppose in the shortest walk, we go through nodes $u_1'=u,\ldots,u_{k'}',u_1,\ldots,u_k=w$ from $u$ to $w$, go through nodes $v_1=w,\ldots,v_l,v_1',\ldots,v_{l'}'=v$ from $w$ to $v$, where $v_1',\ldots,v_{l'}' \notin \{u_1',\ldots,u_{k'}',u_1,\ldots,u_k\}$, and $v_l=u_1$, like the following graph:
u (u_1,...,u_k)
(u_1',...,u_{k'}')\ /\
u_1(v_l) w The walk (u_1,...,u_k) may overlap with (v_1,...,v_l)
(v_1',...,v_{l'}')/ \/
v (v_1,...,v_l)
Now suppose edges $e_1,\ldots,e_h$ are shown in both the walk $(u_1,...,u_k)$ and the walk $(v_1,\ldots,v_l)$, and the remaining edges in both walks are $f_1,\ldots,f_p$ and $g_1,\ldots,g_q$ respectively. Then the cost of the walk $(u_1,...,u_k=v_1,\ldots,v_l)$ is
\begin{align*}
&\sum_{i=1}^h (W_1(e_i)+W_2(e_i))+\sum_{i=1}^pW_1(f_i)+\sum_{i=1}^qW_1(g_i)\\
\ge{}&\sum_{i=1}^h (W_1(e_i)+W_2(e_i))+\frac{1}{2}\left(\sum_{i=1}^p(W_1(f_i)+W_2(f_i))+\sum_{i=1}^q(W_1(g_i)+W_2(g_i))\right).
\end{align*}
That is, the average cost of the walk $(u_1,\ldots,u_k,\ldots,u_1)$ and the walk $(v_l,\ldots,v_1,\ldots,v_l)$ is no more than that of the walk $(u_1,...,u_k=v_1,\ldots,v_l)$. So the shortest walk must have the form shown in the first figure.
Now you can try every node as $t$ to compute the optimal walk from $u$ to $w$ to $v$. This is a polynomial-time algorithm.
