Determine whether the given set, together with the specified operation of addition and scalar multiplication, is a vector space. All vectors of the form $(x, y)$ with addition & scalar multiplication defined as:
$$(x_1, y_1) + (x_2, y_2) = (y_1 + y_2, x_1 + x_2)\quad \text{and}\quad k(x, y) = (k^2x, k^2y).$$
I'm a bit confused in this question like is it ok to change the positions of ordered pair while proving through Axioms. because while proving Axioms the answer is like $(x_1, y_1) + (x_2, y_2) = (x_1+ x_2, y_1 + y_2)$. Please guide.
You still have not specified what the set we are looking at is. Here, I assume that it is the entirety of $\Bbb R^2$ and that the scalar field is $\Bbb R$ with the usual addition and multiplication.
You seem to have the right idea about most of this, though your presentation of checking the associative property appears off. You want to show whether $\left((x_1,y_1)\oplus (x_2,y_2)\right)\oplus (x_3,y_3)$ is equal or not to $(x_1,y_1)\oplus \left((x_2,y_2)\oplus (x_3,y_3)\right)$
Now... if we were to prove that the associative property holds, we would need to be general and keep all of these values as generic variables. To do this, I recommend taking the first expression and then expanding this all out to what the final tuple would equal. If we were to do this, we would have:
$$(x_1,y_1)\oplus \left((x_2,y_2)\oplus (x_3,y_3)\right) = (x_1,y_1)\oplus (y_2+y_3,x_2+x_3) = (y_1+x_2+x_3,x_1+y_2+y_3)$$
We could do the same for the other:
$$\left((x_1,y_1)\oplus (x_2,y_2)\right)\oplus (x_3,y_3) = (y_1+y_2,x_1+x_2)\oplus (x_3,y_3) = (x_1+x_2+y_3,y_1+y_2+x_3)$$
We see that these are not equal.
Since we suspected from the beginning that this would have failed, we could have been far faster at checking this by just seeing if a particular example which is easy to calculate with would have failed. For example, we could have checked to see if $((1,0)\oplus (0,0))\oplus (0,0)$ is equal to $(1,0)\oplus ((0,0)\oplus (0,0))$
It should have been quick to see that the first simplifies to $(1,0)$ and the second simplifies to $(0,1)$. As such, the associative property fails.
The associative property can be long and tedious to check. For examples like this where we suspect that it will fail something, checking the zero property can be even faster. We want there to exist a "zero" element such that "zero" can be "added" on the left or the right of any element and have it be unchanged. We also know that the "zero" element must satisfy $0\otimes v = 0$ where the zero on the left side is the scalar zero and the zero on the right is the "zero" vector.
Here, we are told that $k\otimes (a,b) = (k^2a,k^2b)$. That is again a very suspicious multiplication, but I'll ignore that for now. Using zero for $k$ we find then the zero vector must be $(0,0)$. So... is $(0,0)\oplus (a,b) = (a,b)$?
No! It is not. $(0,0)\oplus (a,b) = (b,a)$ is not equal to $(a,b)$ in general. It is only equal in the case that $a=b$ which is not usually the case. This is yet another axiom failed and as such another proof why the operations do not work to define a vector space correctly. We only needed to show at least one property to fail to be done.
Now... back to the topic of what the set was. If the set was not all of $\Bbb R^2$ but instead only those elements of $\Bbb R^2$ where the first and second entry happened to be equal, then neither of these are valid proofs of the properties failing. Why? Because by replacing all $y$'s with $x$'s (and leaving $x$'s as $x$'s) we see that both expressions are equal in the associativity argument. Same with the zero argument.
For this, we can still show that the scalar multiplication is wrong...
We should have had $2\otimes (a,b) = (1+1)\otimes (a,b) = (1\otimes (a,b))\oplus (1\otimes (a,b))$... but you would have seen the first to be equal to $(4a,4b)$ while the second after all is said and done to be $(2b,2a)$. Even with $a=b$ you would have $2\otimes (1,1)$ by the one line of logic to be equal to $(4,4)$ but with the other line of logic to be equal to $(2,2)$, a contradiction.