The Wolfram Alpha example suggests you want to treat $0$ and $1$ as the Boolean values False and True respectively and, with this convention, to parameterize an arbitrary binary Boolean operator $f$ by means of its truth table values $(a,b,c,d)$:
f[x_, y_, {a_, b_, c_, d_}] := {1 - x, x} . {{a, b}, {c, d}} . {1 - y, y}
(This method exhibits such binary operators as bilinear forms over the field of two elements.)
Here is a comprehensive test proving this works correctly for inputs in the intended domain (that is, all of $a, \ldots, d$ are either $0$ or $1$ and so are $x$ and $y$). It applies all four possible inputs $(x,y)$ and lists the inputs followed by the values of $f$ when applied to them:
Flatten[Array[{#1, #2, f[#1, #2, {a, b, c, d}]} &, {2, 2}, {0, 0}], 1]
{{0, 0, a}, {0, 1, b}, {1, 0, c}, {1, 1, d}}
The second part of the question could be interpreted as asking how the expression
$$f(x,y,(a,b,c,d)) \text{ and } f(y, x \text{ xor } y, (a,b,c,d))$$
(which evidently is a binary Boolean operator in the arguments $x$ and $y$) ought to be parameterized. Let's work it out in steps using the definitions:
and[x_, y_] := f[x, y, {0, 0, 0, 1}];
xor[x_, y_] := f[x, y, {0, 1, 1, 0}];
p = Table[f[x, y, {a, b, c, d}]~and~f[y, x~xor~y, {a, b, c, d}] , {x, 0, 1}, {y, 0, 1}] // Flatten
{a^2, b d, b c, c d}
More generally, if the two occurrences of $f$ are intended to have two different sets of parameters, the same method works:
p = Table[f[x, y, {a, b, c, d}]~and~f[y, x~xor~y, {a1, b1, c1, d1}],
{x, 0, 1}, {y, 0, 1}] // Flatten
{a a1, b d1, b1 c, c1 d}
In other words, a simplified version of the compound input expression is
f[x, y, {a a1, b d1, b1 c, c1 d}]