How to prove that $\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4)$?

Usually, the Dedekind eta function is understood as a function of $\tau\in\mathbb{H}$ (complex upper half plane) where $q=\mathrm{e}^{2\pi\mathrm{i}\tau}$. This way, the modular symmetries of $\eta$ can be expressed easily and the branching issues with $q^{1/24}$ can be avoided. In this answer, I will not make use of modular symmetries, but perhaps some other answer will, so let us stick to the convention of regarding $\eta$ as a function of $\tau$ and use another symbol when we regard it as a function of $q$. Following Michael Somos's example at the product identity website linked in the question, I will write $h(q)$ for $\eta(\tau)$.

Dividing by $h^4(q^4)\,h^4(q^2)$ and isolating the rightmost summand, your identity in question becomes $$\frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)} - \frac{4\,h^4(q^8)}{h^2(q^4)} = \frac{h^4(q)}{h^2(q^2)} \tag{*}$$

Apart from having some power of $q$ as overall multiplier, the factors of eta products (and quotients) can be written as $1+a_n$ with $a_n=\operatorname{O}(q^{kn})$ for some $k>0$ with $k$ and the implied $\operatorname{O}$ constant not depending on $n$, therefore $\sum_{n=1}^\infty |a_n|$ converges absolutely for $|q|<1$, and we can reorder the factors accordingly. Using rules like $$\begin{aligned} \prod_{n=1}^\infty (1-x^{2n}w)(1-x^{2n-1}w) &= \prod_{n=1}^\infty (1-x^n w) \\ \prod_{n=1}^\infty (1+x^n)(1-x^{2n-1}) &= 1 \end{aligned}$$ for $x=q^k$ and applying Jacobi's triple product identity, we find $$\begin{aligned} \frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)} &= \prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m-2})^4 = \left(\sum_{n\in\mathbb{Z}} q^{2n^2}\right)^2 = \vartheta_3^2(q^2) \\ \frac{4\,h^4(q^8)}{h^2(q^4)} &= q\prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m})^2\,(1+q^{4m-4})^2 = q\left(\sum_{n\in\mathbb{Z}} q^{2n(n+1)}\right)^2 = \vartheta_2^2(q^2) \\ \frac{h^4(q)}{h^2(q^2)} &= \prod_{m=1}^\infty (1-q^{2m})^2\,(1-q^{2m-1})^4 = \left(\sum_{n\in\mathbb{Z}} (-q)^{n^2}\right)^2 = \vartheta_4^2(q) \end{aligned}$$ where $\vartheta_2, \vartheta_3, \vartheta_4$ are Jacobi thetanull functions. Therefore (*) is equivalent to the identity $$\vartheta_3^2(q^2) - \vartheta_2^2(q^2) = \vartheta_4^2(q) \tag{**}$$ This is one of six formulae that relate theta nullvalues with those corresponding to halved or doubled period ratios. It can be shown by reordering the nested sum for $\vartheta_4^2(q)$ in a chessboard-like manner: $$\begin{aligned} \vartheta_4^2(q) &= \sum_{u,v\in\mathbb{Z}} (-q)^{u^2+v^2} = \sum_{\substack{r,s\in\mathbb{Z}\\r+s\text{ even}}} q^{r^2+s^2} - \sum_{\substack{r,t\in\mathbb{Z}\\r+t\text{ odd}}} q^{r^2+t^2} \\ &\stackrel{\substack{r = k - m\\s = k + m\\t = k + m + 1}}{=} \sum_{k,m\in\mathbb{Z}} q^{2(k^2+m^2)} - \sum_{k,m\in\mathbb{Z}} q^{\left((2k+1)^2+(2m+1)^2\right)/2} = \vartheta_3^2(q^2) - \vartheta_2^2(q^2) \end{aligned}$$ You may be interested in writing down the other period-ratio doubling and halving formulae and transforming them to eta product identities that involve power products of $h(q), h(q^2), h(q^4), h(q^8)$ only.

I will give another answer here. My idea is to use the dimension formula for a space of modular forms with given weight and level.

First, to find the weight and level of the left-hand side, I will use a result of Gordon, Hughes, and Newman for the modularity of some eta-quotients. I do not write down the statement here. You can find this result in Ken Ono's book 'The Web of Modularity' (cf. Theorem 1.64). According to this result, the eta-quotient in the left-hand side is a modular form of weight 5 for $\Gamma_{0}(8)$ with character $\chi$, where $$\chi(n):=\begin{cases} 1 &\text{if}~n\equiv 1\mod{4},\\ -1 &\text{if}~n\equiv 3\mod{4}. \end{cases}$$

Note that there is a simple formula for vanishing order of an eta-quotient at cusps (cf. Theorem 1.65 of 'The Web of Modularity'). Using this formula, we see that the left-hand side is a cusp form. We denote by $S_{5}(\Gamma_{0}(8),\chi)$ the space of cusp forms of weight 5 for $\Gamma_{0}(8)$ with character $\chi$ as usual.

Using the dimension formula given by Cohen and Oesterle, we obtain $\dim_{\mathbb{C}}S_{5}(\Gamma_{0}(8),\chi)=2$. From theorems 1.64 and 1.65 in Ken Ono's book, we have that $F_{1}:=\eta^{4}(q^{2})\eta^{2}(q^{4})\eta^{4}(q^{8})$ and $F_{2}:=\eta^{4}(q)\eta^{2}(q^{2})\eta^{4}(q^{4})$ are contained in $S_{5}(\Gamma_{0}(8),\chi)$. Note that $F_{1}$ and $F_{2}$ have Fourier expansions starting with $q^{2}$ and $q$, respectively. This follows easily from the infinite product expansion of $\eta$. Thus, $F_{1}$ and $F_{2}$ are linearly independent, hence forms a basis for $S_{5}(\Gamma_{0}(8),\chi)$. By comparing Fourier coefficients (we can compute a finite number of Fourier coefficients of an eta-quotients using Sage), we have $$\frac{\eta^{14}(q^{4})}{\eta^{4}(q^{8})}=4F_{1}+F_{2}.$$