Separating/recovering base signal from two mixed signals, given phase information
Here's the best method I've found so far:
First, I shifted $B_2(x)$ by $\phi$; this shift aligns the underlying (but unknown) signals $w(x)$ and $w(x + \phi)$. Note that once $B_2(x)$ is shifted, the signals will not begin and end at the same $x$-values. I chopped off the values of each signal that didn't overlap. Parts (a) and (b) of the figure below show the original measured signals $B_1(x)$ and $B_2(x)$, and then the aligned signals, which I'll call $s_1(x)$ and $s_2(x)$.
The aligned signals $s_1(x)$ and $s_2(x)$ are not identical, because the base signal $b(x)$ has been added to them at a known phase shift. Therefore, finding the difference between $s_1(x)$ and $s_2(x)$ at each point, and then dividing by the phase shift $\phi$, will give us the numerical derivative of the base signal, $\frac{\triangle b}{\triangle x}$. In summary, the numerical derivative of the base signal is given by
\begin{equation}
 \frac{\triangle b}{\triangle x} = \frac{s_2(x)-s_1(x)}{\phi}.
\end{equation}
Then, to find the estimated base signal $b_{est}(x)$, we numerically integrate $\frac{s_2(x)-s_1(x)}{\phi}$. I corrected for integration drift by numerically integrating forwards and backwards, then averaging the two results.
Lastly, I shifted $b_{est}(x)$ forward by $\frac{\phi}{2}$. I think the reason for a shift of this value is because I computed the numerical derivative between points separated by a full phase, so the derivative approximate should be centered at half of that.
The results (base signal $b(x)$, estimated base signal $b_{est}(x)$, and error between the two) are shown in part (c) of the figure below.