Given $f(x)$, a real function, expressed as the sum of two Gaussian functions:
$$f(x)=\frac{\alpha}{\sqrt{2\pi\sigma_1^2}}\,\,e^{-\frac{(x-\mu_1)^2}{2\sigma_1^2}}+\frac{1-\alpha}{\sqrt{2\pi\sigma_2^2}}\,\,e^{-\frac{(x-\mu_2)^2}{2\sigma_2^2}}$$
where $\alpha \in [0,1]$.
I would like to scale $f(x)$ (i.e. $\hat{f(x)}$), in such a way to obtain zero mean and unit variance, i.e.:
$$\int_{-\infty}^{+\infty}\,\, x\,\hat{f(x)}\,dx=0$$
$$\int_{-\infty}^{+\infty}\,\, x^2\,\hat{f(x)}\,dx=1$$
I will try to provide some additional background and context for the solution of this problem. To correctly answer this question, it is necessary to distinguish between 1) the case of a random variable whose value is the sum of the values of two underlying random variables, and 2) a random variable whose distribution function or density is the sum of two underlying distributions (thus yielding a so-called "mixture" distribution). In the first case it is known that, if the underlying variables are normally distributed and independent, then the sum variable has a normal distribution as well, with its mean being the sum of the two means, and its variance being the sum of the two variances. In contrast, in the second case, generally the mixture distribution resulting from the sum of two different normal distributions is considerably different from a normal one. Since it is obtained as a mixture of two normal distributions with different means, it will commonly have two peaks, provided that the two means are far enough apart. As an example, you can see here how WA plots the bimodal distribution given by the sum of the pdf of $N(20,5) $ and $N(100,10)$.
The OP deals with this second case, as the target function $f (x) $ is the the sum of two normal distributions $f_1 (x) $ and $f_2 (x) $, multiplied by a factor $\alpha $ and $1-\alpha $, respectively. Let us begin by analyzing the simple sum $f_1 (x) +f_2 (x) $, i.e. not considering for now the factors $\alpha$ and $(1-\alpha)$. Let us call $\mu_1$ and $\sigma_1^2$ the mean and variance of $f_1 (x) $, and $\mu_2$ and $\sigma_2^2$ the mean and variance of $f_2 (x) $, respectively. Also, let us assume - without loss of generality - that $\mu_1 \leq \mu_2$. Taking into account the considerations above, we have to shift $f_1 (x) +f_2 (x) $ on the $x $-axis to obtain a function $\hat { f_1 (x) } + \hat { f_2 (x) } $ that is centered to a zero mean, as defined by
$$\int_{-\infty}^{+\infty} x\, [\hat{f_1 (x)} +\hat{f_2(x)}] \, dx=0$$
To do this, we can firstly note that the values of the definite integrals of $x f_1 (x) $ and $x f_2 (x)$, calculated between $-\infty $ and $ \infty $, are equal to the corrisponding mean values $\mu_1$ and $\mu_2$, respectively. This directly results from the definition of the mean value of a continuous distribution. We can also demonstrate this by observing that the indefinite integral of $x \, g(x) $, where $g (x)$ is the pdf of a given normal function with mean $m $ and variance $s^2$, is
$$-\frac {1}{2} m \,\, \text {erf} \left(\frac {m-x}{\sqrt{2} \,\, s}\right ) - \frac {s \,\, \exp{[(m-x)^2/(2s)^2]}}{\sqrt{2 \pi}}$$
where $\text {erf}$ indicates the error function. Calculating this integral between $-\infty $ and $\infty$, the second term reduces to zero, while the first one reduces to $m$ (this depends on the fact that the limits of the given $\text {erf}$ function for $x \rightarrow \infty $ and for $x \rightarrow -\infty $ are $-1$ and $1$, respectively).
So, if we want the function $\hat {f_1 (x)}+ \hat {f_2 (x)}$ to have zero mean, we must have (using the $\hat{} $ symbol to denote not only the translated functions, but also their means):
$$\int_{-\infty}^{+\infty} x \hat {f_1 (x)}+\int_{-\infty}^{+\infty} x \hat {f_2 (x)}=\hat {\mu_1}+\hat {\mu_2}=0$$
Therefore, we have to shift $ f_1 (x) +f_2 (x)$ on the $x $-axis so that $\hat {\mu_1}=-\hat {\mu_2}$. This can be obtained by a translation of magnitude $\displaystyle r=\frac {(\mu_1 +\mu_2)}{2}$, i.e. by substituting $x $ with $X+r$, which leads to a leftward shift of $f_1 (x) +f_2 (x)$ if $r $ is positive and to a rightward shift if $r$ is negative. The new means obtained by this transformation are $\displaystyle \hat {\mu_1}= - \frac {(\mu_2-\mu_1)}{2}$ and $\displaystyle \hat {\mu_2}= \frac {(\mu_2-\mu_1)}{2}$, so that the new overall mean is zero.
By similar considerations we also obtain that, if the two initial underlying functions $f_1 (x) $ and $f_2 (x)$ are multiplied by factors $\alpha$ and $1-\alpha$, we can transform their sum $f (x)$ to another function $\hat {f (x)} $ with zero mean by a shift on the $x $-axis that is equal to the weighted average $r=\alpha \, \mu_1 + (1-\alpha) \, \mu_2$. Taking an example similar to that used above, if $f (x) $ is the sum of the pdf of $N(20,5) $ and $N(100,10)$, but where the former is multiplied by $0.8$ and the latter is multiplied by $0.2$, we get that the zero mean is obtained by a leftward shift on the $x$-axis of $0.8 \cdot 20 + 0.2 \cdot 100=36$, i.e. by substituting $x $ with $X+36$ in $f(x)$. Accordingly, after this transformation, the integral of $x \, \hat {f (x)} $ calculated between $-\infty$ and $+\infty$ (i.e. the mean) is zero, as confirmed by WA here.
Now that $\hat {f (x)}$ is centered to a zero mean, we can scale it to have variance equal to $1$, as defined by
$$\int_{-\infty}^{+\infty} x^2 \hat{f(x)} dx=1$$
To do this, let us note that, considering firstly the translated underlying function $\hat {f_1 (x)} $, we have
$$\int_{-\infty}^{+\infty} x^2 \hat{f_1(x)} dx=\int_{-\infty}^{+\infty} (x- \hat {\mu_1})^2 \hat{f_1(x)} dx + 2 \, \hat {\mu_1} \int_{-\infty}^{+\infty} x \hat{f_1(x)} dx - \hat {\mu_1}^2 \int_{-\infty}^{+\infty} \hat{f_1(x)} dx$$
In the RHS, the first term corresponds to the definition of the variance of $\hat{f_1(x)}$, which is $\sigma_1^2$ (the variance has not changed and is the same of the original function $f_1(x)$). The second term is equal to $2 \, \hat {\mu_1}^2$ (because the integral of $x \, \hat { f_1 (x)}$, as shown above, gives $\hat {\mu_1}$), and the last term is equal to $ -\hat {\mu_1}^2$ (because the integral of $\hat { f_1 (x)}$ is by definition $1$). Thus, we get
$$\int_{-\infty}^{+\infty} x^2 \hat{f_1(x)} dx=\sigma_1^2 + \hat {\mu_1}^2$$
So, if $\hat {f_1 (x)}$ is multiplied by a factor $\alpha $, we get
$$\int_{-\infty}^{\infty} x^2 \, \alpha \, \hat{f_1(x)} \, dx=\alpha \, (\sigma_1^2 + \hat {\mu_1}^2)$$
Similarly, for the other translated underlying function $\hat {f_2 (x)} $, if multiplied by a factor $1-\alpha $, we get
$$\int_{-\infty}^{+\infty} x^2 \, (1-\alpha) \, \hat{f_2(x)} \, dx=(1-\alpha) \, (\sigma_2^2 + \hat {\mu_2}^2)$$
Thus, for the translated sum function $\hat{f (x)}$, we have
$$\int_{-\infty}^{+\infty} x^2 \, \hat{f(x)} \, dx=\alpha \, (\sigma_1^2 + \hat {\mu_1}^2) + (1-\alpha) \, (\sigma_2^2 + \hat { \mu_2 }^2)$$
If we call $t$ this last quantity, we can scale the variance of $\hat {f (x)} $ to $1$ by adding to the function a factor $1/t$. For example, let us take again the function where $f (x) $ is the sum of the pdf of $N(20,5) $ and $N(100,10)$, the first one multiplied to $0.8$ and the second multiplied by $0.2$. The above mentioned substitution $x=X + 36$ shifts the bimodal curve leftward to create a function $\hat {f (x)} $ with zero mean, and where the two translated underlying functions have mean $-16$ and $64$, respectively. The resulting integral of $x^2 \hat {f (x)} $, calculated between $-\infty $ and $+\infty $, has value $0.8 \, [5^2+(-16)^2] + 0.2 \, (64^2+10^2)=1064$. Inserting $1/1064$ as a scaling factor in $\hat {f (x)}$ we get a scaled function with zero mean and unit variance, as confirmed by WA here.
Subtract the mean $\mu=\alpha\mu_1+(1-\alpha)\mu_2$ and divide by the standard deviation, $\sigma=\sqrt{\alpha((\mu_1-\mu)^2+\sigma_1^2)+(1-\alpha)((\mu_2-\mu)^2+\sigma_2^2)}$.