This may be resolved by identifying $B_4$ with the mapping class group of the 4-punctured disk (fixing the boundary). I.e. $B_4\cong Mod(D^2-\{p_1,p_2,p_3,p_4\})$. I will consider isotopy classes of arcs in $D^2-\{p_1,p_2,p_3,p_4\}$ with endpoints on $\partial D^2$ and how they intersect each other, and the action of $B_4$ on the isotopy classes of arcs.
In your pictures defining the generators of $B_4$, I'm assuming the points $p_i$ are numbered in order from top to bottom. Choices of isotopy classes of arcs may be made by choosing a complete hyperbolic metric on the punctured disk so that the boundary is totally geodesic, and making the arcs geodesics. Then the intersection number between pairs of arcs will be minimized by these representatives. Given an arc $w$ and $\sigma \in B_4$, I'll abuse notation $\sigma(w)$ to mean the arc made geodesic after applying $\sigma$ to the arc $w$.
Consider the arcs $x, y, z$:
Then we have that $stab(y)=\langle a,c\rangle$, $stab(x)=\langle b,c\rangle$, $stab(x\cup y)=\langle c\rangle$, which may be seen again by regarding the braid group
as a mapping class group.
What you would like to know is the intersection of the double
cosets $$U=( \langle a,c\rangle\cdot\langle b,c\rangle ) \cap ( \langle b,c\rangle\cdot\langle a,c\rangle ).$$
Suppose that we have $\gamma \in U$. Then one may see that $\gamma(x)\cap y =\emptyset$ and $\gamma(y)\cap x = \emptyset$. Moreover $\gamma(x)\cap \gamma(y)=\emptyset$ since $x\cap y=\emptyset$. Write $\gamma=\alpha \beta, \alpha\in \langle a,c\rangle, \beta \in \langle b,c\rangle$. Then $\beta(x)=x$ and $\alpha(y)=y$, so we see that $\gamma(x)\cap y =\alpha\beta(x)\cap \alpha(y) =\alpha(x)\cap \alpha(y)=\emptyset$, and similar for the other intersection.
Consider the subsurface $P\subset D^2$ obtained by cutting $D^2$ along $x \cup y$, and keeping the middle piece containing $p_2, p_3$. Then $P'=P-\{p_2,p_3\}$ is a twice-punctured disk, and there is only one isotopy class of essential arc with endpoints on $\partial P'$. Then each component of $\gamma(x)\cap P', \gamma(y)\cap P'$ must be isotopic to this arc (the case when one of the arcs is boundary parallel may be dealt with in a similar fashion). However, all but one component of $\gamma(x)\cap P'$ must have endpoints on $x\subset \partial P'$, and all but one component of $\gamma(y) \cap P'$ must have endpoints on $y\subset P'$. Up to composing $\gamma$ with an element of $\langle c\rangle = Mod(P')$, we must see a picture like this:
Thus replace $\gamma$ by $c^{-i}\gamma$ if necessary to
get this configuration. We see now that $\gamma(x)\cap z=\emptyset$,
$\gamma(y)\cap z=\emptyset$. Now focus on the subsurface
$L\subset D^2$ obtained by cutting along $z$ and taking the
left piece containing $p_1, p_2$. There is only one isotopy
class of arc $x$ and $\gamma(x)$ in $L'=L-\{p_1,p_2\}$. But
the endpoints of these arcs lie in $\partial L' -z$, and hence
we see a picture like this:
Thus we see that up to composing with a power of $a$ (which
generates $Mod(L')$), we may assume that $\gamma(x)=x$.
Similarly, up to composing with a power of $b$, we may assume
that $\gamma(y)=y$.
But now $stab(x \cup y) =\langle c\rangle$ in $B_4$. So
we've shown that $\gamma = c^i a^j b^k c^l$.
Now we have $$\gamma = c^ia^jb^kc^l = \alpha \beta, \alpha\in\langle a,c\rangle, \beta\in \langle b,c\rangle.$$
Then $$\alpha^{-1} c^i a^j = \beta c^{-l} b^{-k} \in \langle a,c\rangle\cap \langle b,c\rangle = stab(x\cup y)=\langle c\rangle.$$
Thus $\alpha = c^i a^j c^{-m}, \beta =c^m b^k c^l$, as desired.
,
$b=\ $
and $c=\ $
