There is $f$ such that $\int_0^1 f dx = \lim_{c \downarrow 0} \int_c^1 fdx$, but for $|f|$ this limit does not exist. How is that possible?

There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $\lim_{c \to 0}\int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $\int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $\int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.

The implication $f$ is integrable $\Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.