you can use any numbering system you want as long as you are consistent and bookkeep properly. For instance, coordinate and displacement fields on this quad you have shown are approximated with a summation of "N" shape functions:
x=Sum(X(i)*N(i),i=1,N), same with y, substitute Y(i) for X(i). Say for instance, the coordinates of the 4 nodes are (0,0)=node 1 (lower left), (1,0)=node 2 (lower right) (1,1)=node 3 (upper right), (0,1)=node 4 (upper left). The mapping that describes all 'x' coordinates in this element is x=X(1)*N(1)+X(2)*N(2)+X(3)*N(3)+X(4)*N(4)=
0*N(1)+1*N(2)+1*N(3)+0*N(4)=N(2)+N(3). Similarly, y=N(3)+N(4). More on what N(i) are below.
Now the temperature field (u)
u=Sum(U(i)*N(i),i=1,N)/ U(i) are what you are solving for (you've called it temperature, and labeled them x1,x2,x3,x4).
In isoparametric elements, those N(i) are constructed so that they are 1.0 at the nodes, and 0 at all other nodes (but non zero elsewhere). Say you are using linear elements, and use the numbering system you show (1 at lower left, 2 lower right, 3 upper right, 4 upper left), then N(1)=0.25*(1-eta)*(1-xi),N(2)=0.25*(1+eta)*(1-xi),
N(3)=0.25*(1+eta)*(1+xi),N(4)=0.25*(1-eta)*(1+xi).
(eta,xi) are on the basis element system..every element in your real (x,y) space is mapped to the basis element system, hence the talk about Jacobians above.
What if you didn't like the numbering system? As you suggest, switch 2 and 3. Making sure that you remember that the shape functions need to be 1.0 at the corresponding nodes, and 0.0 at all other nodes, renumber the shape functions:
N(1)=0.25*(1-eta)*(1-xi),N(3)=0.25*(1+eta)*(1-xi),
N(2)=0.25*(1+eta)*(1+xi),N(4)=0.25*(1-eta)*(1+xi).
Remembering that you switched '2' and '3', then X(1)=0,Y(1)=0, X(2)=1,Y(2)=1, X(3)=1,Y(3)=0, X(4)=1,Y(4)=1. The mapping of the 'x' coordinate in this new element becomes:
x=Sum(X(i)*N(i),i=1,N)=X(1)*N(1)+X(2)*N(2)+X(3)*N(3)
+X(4)*N(4)=0*N(1)+1*N(2)+1*N(3)+0*N(4)=N(2)+N(3), same as before, but watch what happens to the mapping of the 'y' coordinate!
y=Y(1)*N(1)+Y(2)*N(2)+Y(3)*N(3)+Y(4)*N(4)=
0*N(1)+1*N(2)+0*N(3)+1*N(4)=N(2)+N(4) (note difference from previous numbering system, y=N(3)+N(4)).
Both numbering systems are perfectly equivalent, as long as you keep the bookkeeping straight! You don't even have to start in the lower left, you could have started the numbering in the upper right, and clockwise, node 1 is upper right, node 2 lower right, node 3 lower left, node 4 is upper left. Now you have to remember to renumber the shape functions N(i):
N(3)=0.25*(1-eta)*(1-xi),N(2)=0.25*(1+eta)*(1-xi),
N(1)=0.25*(1+eta)*(1+xi),N(4)=0.25*(1-eta)*(1+xi).