This answer is to record a sequence of upper bounds which I have been building. I am coming to believe that the correct answer is $2/3$, but I don't have a construction which achieves it.

Looking at the line segment from $(1,0,0,0)$ to $(0,1/2,0,1/2)$, at a point of the form $(u,v,0,v)$ with $u+2v=1$, the function $f$ is at most $u+v$. The same is true for points of the form $(v,u,v,0)$, $(0,v,u,v)$ and $(v,0,v,u)$.

Look at the convex hull of $(0,0,a,b)/(a+b)$, $(a,0,a,b)/(2a+b)$ and $(0,b,a,b)/(a+2b)$. At the first vertex, $f \leq 1$. From the previous paragraph, $f$ is bounded by $(a+b)/(2a+b)$ and $(a+b)/(a+2b)$ at the second and third vertices. In short, at all three vertices of this triangle, $f(w,x,y,z) \leq y+z$. Since this triangle is contained in the plane $by=az$, which contains $(\ast, \ast, 0,0)$, we conclude that we have the bound $f(w,x,y,z) \leq y+z$ throughout this triangle.

After some algebra, one works out that, whenever $wz+xy \leq yz$ we have the bound $f(w,x,y,z) \leq y+z$.

Symmetrically, if $wz+xy \leq wx$, then $f(w,x,y,z) \leq w+x$.

That was the pain-free part.

Let's restrict our attention to the plane $qy = pz$, where $p+q=1$. This meets $\Delta$ in a triangle whose corners are $(0,0,p,q)$, $(1,0,0,0)$ and $(0,1,0,0)$. We can parameterize it as $(w,x,p(1-w-x), q(1-w-x))$, subject to the inequalities $w \geq 0$, $x \geq 0$, $w+x \leq 1$.

The inequality $wz+xy \leq yz$ turns into $(1-p^2) w + (1-q^2) x \leq pq$, a line cutting off one corner of the triangle. Let $r$ and $s$ be the end points of this line segment. The inequality $wz+xy \leq wx$ turns into a hyperbola, passing through the two corners $(w,x)=(1,0)$ and $(w,x)=(0,1)$ and through the centroid $(w,x) = (1/3, 1/3)$. The equation in $w$ and $x$ coordinates should be something like $px(1-2w-x) + qw(1-w-2x)=0$. Let $C$ be portion of the hyperbola which lies inside the triangle.

So, on two regions of the triangle we have bounds. Those bounds then imply, by convexity, bounds on the rest of the triangle. Here is what those bounds are.

Let $t$ be the point on $C$ where $(w+x-y-z)/(wz+xy-yz)$ is minimized.

On $\mathrm{Hull}(r,s,t)$, we know that $f$ is bounded by the linear function which is equal to $y+z$ on the line $rs$ and equal to $w+x$ at $t$. Of course, this would be true for any $t$ on the curve $C$, but we get the best bounds by choosing $t$ as above.
I have not worked out explicit formulas for this, but I will report that, in the special case $(p,q) = (1/2, 1/2)$, it gives the upper bound of $2/3$ through the triangle
$$\mathrm{Hull}( (0,1/3,1/3,1/3,), (1/3,0,1/3,1/3), (1/3,1/3,1/6,1/6) ),$$
which includes the point $(1/4, 1/4, 1/4, 1/4)$.

We have so far given bounds on the triangle $\mathrm{Hull}( (0,0,p,q), r, s)$; on the triangle $\mathrm{Hull}(r,s,t)$, and on one side of the hyperbola $C$. This leaves two regions left over. One is bounded by a portion of $C$, the line segment $rt$, and the line segment from $r$ to $(1,0,0,0)$. For $(w,x,y,z)$ in this region, one gets the best bound by drawing the line segment from $r$ to $(w,x,y,z)$, extending it until it meets $C$, and bounding $f$ by the function which linearly interpolates between the known bounds at $r$ and at $C$.

Similarly, on the region which is bounded by the line segment $st$, the line segment from $s$ to $(0,1,0,0)$ and a portion of $C$, the best bound is interpolating linearly on each line segment from $s$ to $C$.

Now, we could play the game again. We just got bounds throughout the simplex, using planes which contain $(\ast, \ast, 0,0)$. We could restrict those bounds to triangles containing $(0,0,\ast, \ast)$. The resulting function probably would not be convex, so we could take its lower convex hull, getting better bounds. But this computation was too painful for me to want to attempt it.

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