Is finding a single digit of a computation is as hard as finding the computation?
For some functions $f$ you might be correct. For example, we can arrange that $f(|n|) \neq n$ for all $n$ (by controlling the length of $f(x)$) while keeping $f$ arbitrarily difficult, and so $L_1$ reduces to $L_2$ trivially.

Here is a function $f$ satisfying your constraint, for which $L_1$ does not polynomially reduce to $L_2$. The construction uses diagonalization. (There is some ambiguity in your notation, but the proof can be adapted however you resolve the ambiguities.)

Let $L$ be some computable language which requires time $2^{2^{2^n}}$ (such languages exist due to the time hierarchy theorem). We define the function $f$ as follows: $f(\epsilon) = f(0) = 0$, and $f(0x)$ is the indicator of $x \in L$. We define the rest of the function $f$ by giving an algorithm which eventually defines all remaining values.

Let $\phi_i$ be an enumeration of all functions computable in polynomial time. We can construct such an enumeration using timed Turing machines.  We go over the functions $\phi_i$ in order. When it's time to process $\phi_i$, let $n \geq 1$ be the smallest value on which $f$ is undefined. Let $m = \phi_i(0^n,2)$. We consider several different cases:


Case 1: $f(|m|)$ is already defined. If the second digit of $f(|m|)$ is $0$, define $f(n) = 1^n$, otherwise define $f(n) = 0^n$.
Case 2: $f(|m|)$ is not defined, and $|m| \neq n$. Define $f(|m|)$ arbitrarily, and proceed as in case 1.
Case 3: $|m| = n$. Set $f(n) = 10$.


In all cases, we ensure that $f(n) \neq 0^n$ iff the first digit of $0^n$ equals the second digit of $f(|m|)$, showing that $\phi_i$ is not a polytime reduction from $L_1$ to $L_2$.