That description of what Sophie Germain did is slightly glib and inaccurate, I believe. I will give an abbreviated, but accurate, to the best of my knowledge, explanation of what happened:

Note that, if you have proven Fermat's Little Theorem for some exponent, then you have also proven it for every multiple of that exponent. Thus, to prove it for all exponents (> 2) up to some limit, it suffices to prove it for all odd prime exponents up to that limit, as well as for exponent 4. Fermat himself gave an argument which worked for exponent 4, so afterwards, one could consider only odd prime exponents.

Note also that, if p is prime, and we have a solution to x^p + y^p = z^p, then out of {x, y, z}, precisely 0, 1, or all 3 are divisible by p (i.e., if any two were divisible by p, then so would be the third). If indeed all 3 are divisible by p, we may accordingly divide all through by p to obtain a smaller solution; thus, if there is any solution, there is a minimal solution where precisely 0 or 1 of {x, y, z} are divisible by p.

So to prove Fermat's Little Theorem for prime exponent p, it suffices to prove both of the following claims for all solutions to x^p + y^p = z^p:

(A) It cannot be the case that precisely 0 of {x, y, z} are divisible by p

(B) It cannot be the case that precisely 1 of {x, y, z} is divisible by p

If this can be done for every odd prime p up to some limit, FLT is established for all exponents up to that limit.

Germain did not do this. She did not have a strategy for proving (B). This prevented her from establishing FLT in full for any exponent.

Germain did manage, however, to discover a strategy for establishing (A) for various p. Specifically, she discovered a sufficient condition for (A) was the existence of another prime t in a certain decidable relation to p. Germain then manually searched for and discovered such t for each prime p < 100.

Why'd she stop there? Because it seemed like a nice place to stop. Nothing special about 100 except the human factor. One could keep going, and indeed, Legendre extended Germain's results to each prime < 197. Germain and Legendre were both aided by certain techniques they had developed in order to quickly identify certain candidate t that could work as the auxiliary primes for particular exponents p; however, these techniques broke down at 197, and while it is possible to find by brute force an auxiliary prime t in the appropriate relation to p = 197, the smallest such t is very large and the scale of the calculation was beyond feasible at the time.

I may still not have gotten the story exactly correct, or noted all the pertinent details, but for more, see https://www.agnesscott.edu/lriddle/women/germain-FLT/SGandFL..., on which I based the above description.

A little more detail, since I bothered to go learn it:

Lemma: If x and y are coprime, then the gcd of x + y and (x^n + y^n)/(x + y) divides n.

Proof: Expand out the polynomial division (note that x + y does indeed divide x^n + y^n), and then divide the result by (x + y) again, observing a remainder of ny^(n - 1). Thus, the gcd of interest is the same as gcd(x + y, ny^(n - 1)). As y^(n - 1) is coprime to x + y (by coprimeness of x and y), this is furthermore the same as gcd(x + y, n), completing the proof.

Lemma: If x^p + y^p + z^p = 0, for odd prime p, with z indivisible by p, then (x + y) and (x^p + y^p)/(x + y) are coprime p-th powers.

Proof: As z is indivisible by p, so is -z^p = x^p + y^p, and thus so is its factor x + y. At this point, invoking the above lemma and the primeness of p, we find that (x + y) and (x^p + y^p)/(x + y) are coprime; as their product is a p-th power (-z^p), we conclude they are furthermore each p-th powers.

Sophie Germain's theorem: Suppose p is an odd prime, t is a prime, and the mod-t exponent-p case of FLT is true (in the sense that there is no solution to x^p + y^p + z^p = 0 (mod t) where x, y, and z are nonzero (mod t)), but the non-modular exponent-p (A) case of FLT fails (in the sense that there IS some solution to x^p + y^p + z^p = 0 in integers, all indivisible by p, which we can assume minimal so that x, y, and z are pairwise coprime). Then p is a p-th power modulo t.

Proof: Invoking the above lemma, we have some a, b, c such that a^p = y + z, b^p = x + z, and c^p = x + y.

Furthermore, since mod-t FLT is true, we have that (precisely) one of x, y, or z is zero mod t; WLOG, let this be x. But as 2x = b^p + c^p - a^p, we can invoke mod-t FLT again to conclude that one of b, c, or a is zero mod t. It cannot be b or c, as y and z, respectively, are nonzero mod t; thus, a is zero mod t. Thus, y + z = 0 (mod t), and therefore, by expanding out the polynomial (y^p + z^p)/(y + z), we see that its integer value is equal to py^(p - 1) modulo t. As x is zero mod t, we must have furthermore that (x^p + y^p)/(x + y) must be y^(p - 1) modulo t (note that this is nonzero modulo t). As the former and latter are both p-th powers by the above lemma, so is their ratio in modulo t arithmetic, which is p, completing the proof.

We can now rephrase Sophie Germain's theorem like so: To establish the (A) case of FLT for exponent p, it suffices to find some prime t such that BOTH the mod-t exponent-p case of FLT is true AND p is not a p-th power modulo t.

Such a t is the auxiliary prime for p discussed above. Note that the truth or falsehood of this condition on t relative to p is decidable by finite search.

A couple more comments: As a consequence of the multiplicative group modulo a prime being cyclic, we have that, for primes p and t, EVERY value is a p-th power modulo t unless t is 1 mod p. Thus, an auxiliary prime t for odd prime p must be of the form kp + 1. Furthermore, as t cannot be 2, it must be an odd prime, and therefore k must be even. Finally, we can also rule out k divisible by 3 by again invoking the cyclicity of the multiplicative group modulo t (were k divisible by 3, we would have some primitive cube root of 1 modulo t which was furthermore a p-th power; the three powers of this value would sum to zero and thus provide a counterexample to the mod-t exponent-p case of FLT). So when searching for auxiliary t to prime p, we are looking for primes of the form kp + 1 where k is an even value not divisible by 3; for all such k up through 16, Germain and Legendre managed to prove that conversely, whenever kp + 1 is prime, it satisfies all the conditions to be an auxiliary prime for p (with a slight exception for the two cases where p = 3 and k is 14 or 16), which provides auxiliary primes for each p < 197; however, at p = 197, one must go beyond this, to a minimal k of 38.

I do not at the moment know if there are auxiliary primes for each p (and I suspect no one does?), nor how one goes about showing that, for various k, kp + 1 is automatically an auxiliary prime whenever prime (or whether this generalizes beyond k = 16). Oh well. More to learn, always.