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This is more an extended comment than an answer.

In my first comment I proposed to rewrite the series $f(x)$ as a series $$\small \begin{eqnarray} \hat g(t)&=& \big ( (1-t )-1 \big) - \big((1-t)^2-1\big) + \big((1-t)^{2^2}-1\big) - ... + ... \\ &=& (-t) - ( -2t+t^2 ) + (-3t+3t^2-t^3 ) -... + ... \end{eqnarray}\tag 1 $$ such that -hopefully- we can write $ \small f(x)= \hat g(1-x) + \frac 12 $ and for example $ \small f(0.99) = \hat g(0.01) + \frac 12 $ by some much smoother evaluatable series.

This ansatz did not solve the problem fully, but gives an interesting detail, which I should possibly ask in a separate question.

I proceeded from eq (1) by formally collecting equal powers of the argument $t$ getting at each power of $t$ an infinite series, which are all divergent, but are compositions of geometric series with alternating sign - and usually we can use the closed forms for this. What I got is the following $$\small \begin{eqnarray} \hat g(t)&=& &(-t)\cdot & ( 1-2+4-8+16-...+...) \\ && + &(-t)^2 \cdot & ( -1 + 6 -28 +120- ... + ...)\\ && + & & ... \\ && + & (-t)^k \cdot & \sum_{j=1}^\infty (-1)^j \binom{2^j}{k} \\ && \vdots \end{eqnarray} \tag 2$$ It is now interesting, that the binomial-series have closed forms: they can exactly be expressed by Stirlingnumbers 1st kind and the closed forms of geometric series with quotients $-(2^k)$ so that for instance the first few coefficients can be written as $$\small \begin{eqnarray} \hat g(t)&=& &(-t)\cdot &\frac 1{1!}& ( + 1{1 \over1+2^1} &&& ) \\ && + &(-t)^2 \cdot &\frac 1{2!}& ( -1{1 \over1+2^1} &+1{1 \over1+2^2}&& ) \\ && + &(-t)^3 \cdot &\frac 1{3!}& (+ 2{1 \over1+2^1} &-3{1 \over1+2^2}&+1{1 \over1+2^3}& ) \\ && \vdots \end{eqnarray} \tag 3$$ If we change order of evaluation now, then we find columnwise the mercatorseries for $\log(1-t)$ and its powers so that we can write $$ \small \hat g(t) = {\log(1-t)\over 1! }{1\over 3} + {\log(1-t)^2\over 2! }{1\over5} + ... \tag 4$$ We can now even include the $ + \frac 12$-term into the function $ \hat g(t)$ so that we have finally $g(t)= \hat g(t)+ \frac 12$ and $$ g(t) = \sum_{k=0}^\infty {\log(1-t)^k\over k! }{1\over 1+2^k} \tag 5$$

But this decomposition into double series and change order of computations is not yet enough to reproduce $f(x)$. My first impression was that this is a road to nirvana, but now I see, that the error has a systematic pattern and thus might possibly be "cured" by one more step of analyis: the curve of the differences $d(x) = f(x)-g(1-x)$
a) has a sinusoidal pattern and -even better -
b) seem to have a constant amplitude.

Here are pictures of the function $f(x)$, $g(1-x)$ and $d(x)= g(1-x)-f(x)$: The blue line and the magenta line are nicely overlaid over the whole range $0<x<1$ and the amplitude of the red error-curve (the $y$-scale is at the right side) seems to be constant.
For the more critical range near $x=1$ I've a zoomed picture:

But from here my expertise is exhausted and I'm missing the tools to proceed. First thing would be to look at the Fourier-decomposition of $d(x)$. Alternatively, possibly we have here something like in the Ramnujan-summation where we have to add some integral to complete the divergent sums, but I really don't know.

But if we have in fact the amplitude of the error (=difference $d(x)$) constant, then a discussion of the (likely) easier series $g(t)$ might allow to infer to the properties of the oscillations in $f(x)$

[update] Crazy. If I give the x-axis a logarithmic scale, the curve of the $d(x)$ looks like a perfect sine-wave. If this is true, then because $g(0)=1/2$ the non-vanishing oscillation of $f(x)$ when $x \to 1$ is obvious because it's just the oscillation of the $d(x)$-curve...