You need to understand two things:
How to translate Turning Turtles to Nim, and back.
What the winning strategy is in Nim.
Turning Turtles $\iff$ Nim
A coin with the Head side up at position $i$ is equivalent to a Nim heap of size $i$. That is, $\mathrm{H\,T\,T\,H\,T\,T\,H\,T\,H\,T}$ is equivalent to a Nim position with heaps of size $1,4,7$ and $9$.
Flipping coins numbered $i$ and $j$, where $i<j$, so that coin number $j$ was originally heads, is equivalent to reducing the Nim heap of size $j$ to a Nim heap of size $i$. For example, from $\mathrm{H\,T\,T\,H\,T\,T\,H\,T\,H\,T}$, you could flip coins number $2$ and $7$, resulting in $\mathrm{H}\,\bf{H}\,\mathrm{T\,H\,T\,T}\,\bf{T}\,\mathrm{T\,H\,T}$. If you translate both to Nim positions, this looks like
$$
(1,4,7,9)\longrightarrow (1,2,4,9),
$$
which is indeed like reducing a Nim heap of size $4$ to one of size $2$.
The exception to this correspondence is when both coins are turned from heads to tails. For example, moving from $\mathrm{H\,H,H}$ to $\mathrm{T\,H\,T}$. In this case, the first Nim position was $(1,2,3)$, and it seems like the result should be $(1,1,2)$, since we flipped coins $3$ and $1$. Here, we have to mentally delete any repeated heaps, so instead of $(1,1,2)$, we just get a single Nim heap of size $2$. These repeated heaps do not affect the Nim outcome.
Now that you know how Turning Turtles positions and moves correspond to Nim, all you need to know is$\dots$
Winning Strategy in Nim
To find a winning move in a Nim position with heap size $n_1,\dots,n_k$, all you need to do is the following:
Compute the Nim sum of the heap sizes: $m=n_1\oplus \dots \oplus n_k$. A winning move exists iff $m\neq 0$.
Compute $n_i\oplus m$ for each $i\in \{1,\dots,k\}$. If $n_i\oplus m$ is a smaller number than $n_i$, then reducing the $n_i$ heap to $n_i\oplus m$ is a winning move.
Example
Let's find all winning moves starting from $\mathrm{H\,T\,T\,H\,T\,T\,H\,T\,H\,T}$. The corresponding Nim position is $(1,4,7,9)$, whose Nim sum is $1\oplus 4\oplus 7\oplus 9=11$. Then,
$11\oplus 1=10$, so there is no winning move on the $1$ heap.
$11\oplus 4=15$, so there is no winning move on the $4$ heap.
$11\oplus 7=12$, so there is no winning move on the $7$ heap.
$11\oplus 9=2$, so reducing the $9$ heap to $2$ is a winning move.
Translating this back to Turning Turtles, the only winning move is to flip over coins $2$ and $9$.
Best Answer
Surreal numbers are just a subclass of the class of Games that happen to have some very interesting algebraic properties. Conway defined and studied them for these algebraic properties and not for their use in playing games, although he first thought of them in this context. In fact, most partizan games humans play have integers and dyadic fractions but no more surreal values.
If your question was instead about the broader concept of having a number-like system for combinatorial games, including much more than surreal numbers, then it is clear that there are many applications to game analysis, including all impartial games [1] and many partizan games like Domineering [1], Fox-and-Geese [1], Hex [2], Go endgames [3] and even some chess endgames [4].
[1] Berlekamp, Conway, Guy. "Winning Ways for your Mathematical Plays"
[2] Selinger. "On the combinatorial value of Hex positions"
[3] Berlekamp, Wolfe. "Mathematical Go: Chilling Gets the Last Point"
[4] Elkies. "On numbers and endgames: Combinatorial game theory in chess endgames"