Are space and time hierarchies even comparable?
You can get the situation you describe by choosing weird functions $f(n)$ and $g(n)$. 

For example, let $g(n) = n^3$ and $$f(n) = \begin{cases}
 n &amp; \text{if $n$ is odd},
\\\
 2^{n^5} &amp; \text{if $n$ is even}.
\end{cases}
$$

Then choose $L_1$ and $L_2$ as follows:

$L_1$ is a language containing only strings of even length which can be decided in time $O(2^{n^5})$ but not in time $O(2^{n^4})$. The existence of such a language is pretty easy to prove from the time hierarchy theorem.

$L_2$ is a language containing only strings of odd length which can be decided in space $O(n^3)$ but not in space $O(n^2)$. The existence of such a language is pretty easy to prove from the space hierarchy theorem.

Then we have the following facts:

$L_1 \in TIME(f(n))$:

To decide whether a string is in $L_1$, simply check whether the length $n$ is even. If it is, then continue to use the $O(2^{n^5})$ time decider for $L_1$ whose existence is guaranteed by the definition of $L_1$. If $n$ is odd, immediately reject since $L_1$ does not include any odd length strings anyway. This procedure decides $L_1$, runs in time $O(n)$ when $n$ is odd, and runs in time $O(2^{n^5})$ when $n$ is even. In other words, this procedure decides $L_1$ in time $O(f(n))$. As desired, $L_1 \in TIME(f(n))$.

$L_2 \in SPACE(g(n))$:

By the definition of $L_2$, $L_2$ can be decided in space $O(n^3)$. Thus, $L_2 \in SPACE(n^3) = SPACE(g(n))$, as desired.

$L_1 \not\in SPACE(g(n))$:

Suppose for the sake of contradiction that $L_1 \in SPACE(g(n)) = SPACE(n^3)$. We know that $SPACE(n^3) \subseteq TIME(2^{O(n^3)}) \subsetneq TIME(2^{n^4})$. Thus, there exists a decider for $L_1$ which runs in time $O(2^{n^4})$. This directly contradicts the definition of $L_1$. Then by contradiction, we see that $L_1 \not\in SPACE(g(n))$.

$L_2 \not\in TIME(f(n))$:

Suppose for the sake of contradiction that $L_2 \in TIME(f(n))$. This means that there exists a constant $c$ and an algorithm $A$ deciding $L_2$ such that on any input of size $n$, algorithm $A$ terminates in time $c\times f(n)$. 

We construct a new algorithm $A'$ as follows: given some input, walk through the entire input, keeping track of whether the input length is even or odd; if at the end of the input the length is determined to be odd, return to the start of the input and run $A$; otherwise, reject. For any input of odd length, $A'$ returns the same answer as $A$. For any input of even length, $A'$ rejects, which matches the expected behavior since $L_2$ contains no even length strings. Thus, $A'$ also decides $L_2$. On even length inputs, $A'$ runs for exactly $n$ steps. On odd length inputs, $A'$ runs for exactly $2n$ steps more than $A$ requires. But $A$ requires at most $c\times f(n)$ steps, which for odd $n$ is $cn$. Thus, in all cases, $A'$ runs in at most $(c+2)n$ steps. In other words, algorithm $A'$ decides $L_2$ in time $O(n)$. 

But since $TIME(n) \subseteq SPACE(n)$, we can conclude that $L_2 \in SPACE(n) \subsetneq SPACE(n^2)$. This contradicts the definition of $L_2$. Thus, by contradiction we see that $L_2 \not\in TIME(f(n))$.