Is this an abuse big O notation as a power of a number?
This is perfectly standard use of big O notation, despite some purists disliking it. Here is what the passage you quoted means:

Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) = O(t(n))$.

This is still somewhat ambiguous, since it’s perhaps not clear what variables the big O depends on. In this case, big O hides a universal constant:

There exists a constant $C&gt;0$ such that the following holds.
Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) \leq Ct(n)$ for all $n$.

This allows us to rephrase the statement in yet another way:

There exists a constant $C&gt;0$ such that the following holds.
Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{Ct(n)}$ time deterministic single tape Turing machine.

(This assumes that by “$f(n)$ time machine” we mean a machine running in time at most $f(n)$.)