I am with trouble to prove this statment:
"Suppose that $\{\beta_1,...,\beta_n\}$ is a base of $L|K$ and $\mathcal{M}$ is a subset of some fied $M\supseteq L$. Prove that $\{\beta_1,...,\beta_n\}$ generates $L(\mathcal{M})|K(\mathcal{M})$."
My idea was that: we know that $$L[\mathcal{M}]=\left\lbrace\sum\limits_{j=1}^s\alpha_j\prod\limits_{i=1}^rm_i:\alpha_j\in L,\,m_i\in\mathcal{M},\, 1\le i\le r,\, 1\le j\le s,\, r,s\in\mathbb{N}\right\rbrace.$$ So, to prove that $\{\beta_1,...,\beta_n\}$ generates $L[\mathcal{M}]$, is suffice prove that generates the elements of form $\alpha\prod\limits_{i=1}^rm_i$, where $\alpha\in L$ and $\{m_1,...,m_r\}\subseteq\mathcal{M}$, $r\in\mathbb{N}$. If $\alpha=a_1\beta_1+...+a_n\beta_n$, exist $b_1,...,b_n\in K$ such that $\alpha^{-1}=b_1\beta_1+...+b_n\beta_n$, and we have \begin{align*} \left(\alpha\prod\limits_{i=1}^rm_i\right)^{-1}=\alpha^{-1}\prod\limits_{i=1}^rm_i^{-1}= (b_1\beta_1+...+b_n\beta_n)\prod\limits_{i=1}^rm_i^{-1}=\sum\limits_{j=1}^n\left(b_j\prod\limits_{i=1}^rm_i^{-1}\right)\beta_j, \end{align*} so, $L[\mathcal{M}]|K(\mathcal{M})\subseteq\langle\beta_1,...,\beta_n\rangle$. On the other hand, we know that $L(\mathcal{M})=Frac(K[\mathcal{M}])$, and if $\alpha,\beta\in\langle\beta_1,...,\beta_n\rangle$, then $\alpha\beta\in\langle\beta_1,...,\beta_n\rangle$. Thus, to prove that $L(\mathcal{M})|K(\mathcal{M})\subseteq\langle\beta_1,...,\beta_n\rangle$, is suffice prove that $\alpha^{-1}\in\langle\beta_1,...,\beta_n\rangle$, for all $\alpha\in L[\mathcal{M}]|K(\mathcal{M})$, $\alpha\ne0$. But I can't prove that any inverse of $L[\mathcal{M}]|K(\mathcal{M})$ belongs to $\langle\beta_1,...,\beta_n\rangle$.
Remarks: $\langle\beta_1,...,\beta_n\rangle$ is the vector space generate by $\{\langle\beta_1,...,\beta_n\}$, $K[\mathcal{M}]$ ($K(\mathcal{M})$) is the ring (field) obtained by adjunction $\mathcal{M}$ to $K$ and $Frac(D)$ is the field of fractions of a domain $D$.
not sure how much help you want, but you are probably just over thinking this a little, or have slightly misunderstood the question (unless I have!). I hope the following will help. In particular, try and stick with working directly with the fields.
You have by definition that $L(\mathcal{M})$ is the smallest subfield of $M$ containing the elements of $L$ and the elements of $\mathcal{M}$, and similarly for $K(\mathcal{M})$, so both are fields, and you should see that this $L(\mathcal{M})/K(\mathcal{M})$ is a field extension (think inclusion).
So, you just need to show that any element of $L(\mathcal{M})$ can be written in the form $\displaystyle\sum_{1 \leqslant i \leqslant n} k_i\cdot \beta_i$, where $k_i \in K(\mathcal{M})$. One approach is to specify a basis of $L(\mathcal{M})$ over $L$, write a general element of $L(\mathcal{M})$ in terms of this basis over $L$, and think about how you can then rewrite this general element of $L(\mathcal{M})$ in the required form.
I wonder by your proposed method if you had the right idea behind the generators of a field extension? It is just linear algebra, and you are looking for a spanning set of the larger field over the base field.