Find a strictly decreasing $f:(0,1) \to [0, \infty]$ such that $f(0^{+})= \infty$, $f(1)=0$ and $f(u^2)=2 f(u)$.

Assume that $$f$$ is a function from $$(0,1]$$ to $$[0, \infty]$$ that is strictly decreasing and satisfies $$f(1)=0$$, $$f(0^{+})= \infty$$ and $$f(u^{2})=2 f(u)$$. I do not know if it helps, but let us assume that $$f$$ is convex. There exists one family of solutions given by the set $$\lbrace x \to -k \ln x, k>0 \rbrace$$. Is there another family of solutions that does not lie in the above set?

Let $$g(x):=f\big(\exp(-x)\big)$$ for $$x\in\mathbb{R}_{>0}$$. First, I shall prove that every solution $$f:(0,1)\to\mathbb{R}$$ to the functional equation $$f(u^2)=2\,f(u)$$ for all $$u\in(0,1)$$ takes the form $$f(u)=-\ln(u)\,h\left(\frac{\ln\big(-\ln(u)\big)}{\ln(2)}\right)\tag{*}$$ for every $$u\in(0,1)$$, where $$h:\mathbb{R}\to\mathbb{R}$$ is a periodic function of period $$1$$.
We have $$g(2x)=f\big(\exp(-2x)\big)=f\Big(\big(\exp(-x)\big)^2\Big)=2\,f\big(\exp(-x)\big)=2\,g(x)$$ for all $$x>0$$.
Let $$h(t):=\dfrac{g\big(2^t\big)}{2^t}$$ for all $$t\in\mathbb{R}$$. Then, $$h(t+1)=\dfrac{g\big(2^{t+1}\big)}{2^{t+1}}=\frac{g\big(2\cdot 2^t\big)}{2\cdot 2^t}=\frac{2\cdot g(2^t)}{2\cdot 2^t}=\frac{g(2^t)}{2^t}=h(t)\,.$$ Therefore, $$h$$ is a periodic function with period $$1$$. Ergo, $$f(u)=g\big(-\ln(u)\big)=-\ln(u)\,h\left(\frac{\ln\big(-\ln(u)\big)}{\ln(2)}\right)$$ for every $$u\in(0,1)$$.
With the additional assumption that $$f$$ is continuous and nonnegative, $$h$$ is also continuous and nonnegative. From the limit conditions on $$f$$, we have $$\lim\limits_{x\to 0^+}\,g(x)=0\text{ and }\lim\limits_{x\to\infty}\,g(x)=\infty\,.$$
This shows that $$h$$ is bounded below above $$0$$, namely, $$h(t)\geq l$$ for every $$t\in\mathbb{R}$$, where $$l>0$$ is fixed. These are all required properties of $$h$$. That is, all solutions $$f$$ with the continuity, nonegativity, and limit conditions must be of the form (*), where $$h:\mathbb{R}\to\mathbb{R}_{>0}$$ is a continuous periodic function with period $$1$$.
In particular, if $$h\equiv k$$ for some constant $$k>0$$, then we get the family of solutions that the OP proposed. While it has not been taken into account that $$f$$ is strictly decreasing, it is an easy exercise to show that there are nonconstant periodic functions $$h$$ that makes $$g$$ a strictly increasing function, which in turns makes $$f$$ a strictly decreasing function. I have not thought about the convexity condition, but it is highly probable that only convex solutions take the form proposed by the OP.
If you only need $$f$$ to be strictly decreasing, then $$h(t):=1+\epsilon\,\sin(2\pi t)$$ for all $$t\in\mathbb{R}$$ works for any real number $$\epsilon$$ such that $$|\epsilon|\leq \dfrac{\ln(2)}{\sqrt{4\pi^2+\big(\ln(2)\big)^2}}\approx 0.10965$$. This gives another family with parameter $$\epsilon$$.