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Of course $L$ can be viewed as a 2-dimensional $K$-vector space, and thus every element $x \in L$ can be written as $x=\mu + \nu \alpha\;$ for some $\mu,\nu \in K, \alpha \in L$. Also, $\text{char}(K)\neq 2$.

Let $\tau: L \rightarrow L$ be given by $\tau(\mu + \nu \alpha) = \mu - \nu\alpha$.

Now, I want to show that $\tau \in \text{Aut}_{K}(L)$.

$\tau(x+y) = \tau(x)+\tau(y)$ is easy, but I can't show $\tau(xy) = \tau(x)\tau(y)$.

Does $\tau \mid_{F}=\text{id}_{F}$ follow directly from the definition of $\tau$ with $\nu = 0$?

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    For that to be an automorphism you need $\alpha^2\in K$. Unless we are in characteristic two you can always find such an element $\alpha\in L\setminus K$, but it does not work for an arbitrarily chose $\alpha$. – 2017-01-19
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    For example, consider $K=\Bbb{R}$, $\alpha=1+i\in\Bbb{C}$. You are claiming that $1+i$ and $-1-i$ should be conjugates. But their minimal polynomials over $\Bbb{R}$ are different. – 2017-01-19
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    @JyrkiLahtonen: Thank you. Yes, the characteristic is not 2. – 2017-01-19

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$$\tau((\mu +\nu\alpha)(x+y\alpha))= \tau(\mu x +\nu y\alpha^2+(\nu x +\mu y)\alpha)=\mu x +\nu y\alpha^2-(\nu x +\mu y)\alpha=\mu (x-y\alpha)-\nu\alpha (x-y\alpha)= (\mu -\nu\alpha)(x-y\alpha)=\tau(\mu +\nu\alpha)\tau(x+y\alpha),$$ because $\alpha^2\in K $ (just a matter of degree).

Indeed $\tau|_K=id_K $ ($F$ in your comment is a typo) is a matter of definition, since your $\mu,\nu\in K $ are unique.

I think you want to calculate the Galois group $Gal(L|K) $. Then you should show $id_L\neq \tau $. Since $|Gal (L|K)|=[L:K]=2$ your group just consists of $id_L,\tau $ and hence is $Gal (L|K)=\mathbb {Z}_2$.

But I think it is easier to use the degree argument of $L|K$ directly. Then you do not have to show $Gal (L|K)=\mathbb {Z}_2$ by calculating its elements.