Take a look at the definition of a polynomial ring. By $R[x]$ we look at all polynomials with coefficients in our ring $R$. Note that its crucial that we not only look at the linear polynomials when "adjoining" $x$, but also at the quadratic, the cubic, etc. So in reality we not only "adjoin" $x$ as an intermediate variable but also $x^2$, and $x^3$, and $x^4$, and so on. Simply said: we look at all powers of our new element.
So at a first glance you can look at $K[\alpha]$ as the polynomial ring over $K$ in the variable $\alpha$ but now we are in the reign of fields and want to have a field extension, saying that adjoining (now I omit the quotation marks as the process is called so) actually yields a field again. So, we not only need $\alpha,\alpha^2,\dots$ but also $\alpha^{-1},\alpha^{-2}\dots$ to regain our field structure. This new field we call $K(\alpha)$, saying we adjoin $\alpha$ to $K$ and also its inverse such that we have a field again. This might shed some light on first question. Also, take a look at this question dealing with the difference between algebraic and transcendental extensions regarding the question of inverses.
Regarding your second question, we are now dealing with a specific field extension. Your notation seems a bit confused so I will just assume that you meant to write $\Bbb Q(\sqrt2,\sqrt5)$. So, first lets start with $\Bbb Q(\sqrt2)$. Let us first think about $\Bbb Q[\sqrt2]$, the polynomial ring over $\Bbb Q$ with $\sqrt2$ as a variable. According to our definition of polynomial rings the elements of $\Bbb Q[\sqrt2]$ are of the form $a_0+a_1\sqrt2+a_2(\sqrt2)^2+a_2(\sqrt2)^3+\cdots+a_n(\sqrt2)^n$ for some positive integer $n$. But now we know in addition $(\sqrt2)^2=2$. So all powers of $(\sqrt2)^k$ can be reduced to either a multiple of $2$ or to a multiple of $2$ times $\sqrt2$. Esssentially all boils down to $\Bbb Q[\sqrt2]=\{a+b\sqrt2~|~a,b\in\Bbb Q\}$, we do not need the rest of the powers to generate all of $\Bbb Q[\sqrt2]$ and this fact is encrypted in the minimal polynomial $f(x)=x^2-2$ over $\Bbb Q$. Thinking for a moment of $\sqrt2$ as just an expression for the solution of $x^2=2$ what we actually do by adjoining $\sqrt2$ is to manually demanding this equation to be solvalbe by adjoining an element, called $\sqrt2$, which is defined to fulfill $f(\sqrt2)=0$. This happens when forming the quotient $\Bbb Q[X]/(x^2-2)$.
It might be helpful to adopt this point of view when thinking about adjoining elements. Regarding $\Bbb Q(\sqrt2)$ we note that $\sqrt2^{-1}=\frac1{\sqrt2}=\frac{\sqrt2}2\in\Bbb Q[\sqrt2]$ and similiar for the powers so we already can conclude $\Bbb Q[\sqrt2]=\Bbb Q(\sqrt2)$. This would not work if we would consider $f(x)=x^2-2$ as a polynomial over $\Bbb Z$ so there we would have to adjoin the inverse of $\sqrt2$ (if needed) separately.
Now, if we have two elements adjoined the situation is a polynomial ring in two variabels $\Bbb Q[\sqrt2,\sqrt5]$ where there elements can have all combinations of powers of $\sqrt2$ and $\sqrt5$. But again, examining the first few we notice that eventually we always get rational multiplies of either $\sqrt2,\sqrt5$ or $\sqrt{10}$. So your ring looks like $\Bbb Q[\sqrt2,\sqrt5]=\{a+b\sqrt2+c\sqrt5+d\sqrt{10}~|~a,b,c,d\in\Bbb Q\}$. Again, inverses follows as in the case of only adjoining $\sqrt2$ so $\Bbb Q[\sqrt2,\sqrt5]=\Bbb Q(\sqrt2,\sqrt5)$. To get a more intuitive understanding you should note that you can view a field extension as a vectors space over the base field of dimension the degree of the extension. $\Bbb Q(\sqrt2,\sqrt5)$ has degree $4$, so the vector space is of dimension $4$ and a basis is given by $\mathfrak B=\{1,\sqrt2,\sqrt5,\sqrt{10}\}$. Similiar $\Bbb Q(\sqrt2)$ is of degree $2$, so the vector space is of dimension $2$ and the extension $\Bbb R/\Bbb Q$ is infinite, so $\Bbb R$ can be viewed an infinite dimensional vector space over $\Bbb Q$.
Regarding your last question, you missed a crucial fact: the minimal polynomial has to be monic, i.e. the leading coefficient has to be $1$. That also answers the part where you were asking about $R[x]/(ax+b)$. First, $ax+b$ is not monic, and second even when looking at $R[x]/(x-a)$ you just get $R$ as $a$ already has to be in $R$ as the polynomial you mod out has to be in $R[x]$. But yes, finding an irreducible, monic polynomial $f(x)\in K[x]$ such that $f(\alpha)=0$ guarantees you that this is the minimal polynomial of $\alpha$ over $K$. One can actually show that the minimal polynomial is unique, and thus finding a polynomial that satisfies all requirements already is the minimal polynomial.
I would do it this way, without any linear algebra, because this is a rather simple situation:
Consider $\alpha$ as an irrationality over the Gaussian numbers, $\Bbb Q(i)$. You know that $X^4-5$ is still irreducible as a $\Bbb Z[i]$-polynomial, because $(5)$, though not a prime, is product of two distinct primes, $5=(2+i)(2-i)$, and Eisenstein applies. That is, $f(X)=\text{Irr}\bigl(\alpha,\Bbb Q(i)[X]\bigr)=X^4-5$ and by translation, $\text{Irr}\bigl(\alpha+i,\Bbb Q(i)[X]\bigr)=f(X-i)=X^4-4iX^3-6X^2+4iX-4=g(X)$. Now your desired polynomial is just $g(X)\overline g(X)$, where the bar is complex conjugation of each coefficient.
Best Answer
There appear to be some typos in your post, so if you update your question I think that this answer will no longer be valid.
Let $K=\mathbb{Q}$ and $E=\mathbb{Q}(\pi)$. Take $\alpha=\pi$ and $\beta=\frac{\pi+1}{\pi-1}$. Note that $2(\beta-1)^{-1}+1=\alpha$, so that $K(\alpha)=K(\beta)$.
However, there cannot be a $\gamma\in K^\times$ such that $\alpha=\gamma^2\beta$ or we would have $$ \pi=\gamma^2\frac{\pi+1}{\pi-1} \quad \Rightarrow \quad \pi^2-(\gamma^2+1)\pi-\gamma^2=0 $$ which is a contradiction, since $\pi$ is transcendental over $\mathbb{Q}$.
I have assumed some of the corrections left as a comment on your post, except that $[E:K]=2$. So if you meant that in addition to the fact that $K$ is not characteristic $2$ this answer obviously fails.
EDIT: In fact, I don't even think that the criterion of $[E:K]=2$ is enough to salvage this question. Consider $K=\mathbb{Q}(\pi^2)$ and $E=\mathbb{Q}(\pi)$. This is a degree $2$ extension, and if all other choices are made as above then you still end up with a contradiction (though the final contradiction takes a bit more effort to show that it is a nonzero polynomial).