The latter factorizes mod 2 as $x(x+1)$, so $(2)=\left(2,\frac{1+\sqrt{−199}}{2}\right) \left(2,\frac{3+\sqrt{−199}}{2}\right)$, neither of the factors being principal (and hence both of order 2)
Why do you think that these prime ideals being nonprincipal implies that they have order 2?
Labeling the prime ideals, say $(2) = \mathfrak{p}\mathfrak{p}'$. If the order of $\mathfrak{p}$ is 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$ (hint: you can use that they have the same norm and are equal in the class group). Can you show that isn't the case?
In fact, $\mathfrak{p}^k$ isn't principal for $k = 1,2,\dots,8$, but $\mathfrak{p}^9 = \left(\frac{43\pm\sqrt{-199}}{2}\right)$ (sign depending on choice of $\mathfrak{p}$). Hence the order of $\mathfrak{p}$ in the class group is 9.
Well, he is correct, there is an isomorphism with the group of equivalence classes of positive binary forms of discriminant $-199,$ under Gauss composition...
The easiest way is to find equivalent forms which all have the same middle term, so that Dirichlet's description of Gauss composition is evident. In this case, forcing the middle coefficient to be $43$ works nicely. Each triple $\langle a,b,c \rangle$ refers to the binary form
$$ f(x,y) = a x^2 + b xy + c y^2. $$
From Henri Cohen, A Course in Computational Number Theory, especially pages 225-229: we have Theorem 5.2.8, when $D<0$ is congruent to $0$ or $1 \pmod 4,$ we have a mapping from the positive forms (well, eqivalence classes) of that discriminant $$ \langle a,b,c \rangle \; \; \mapsto \; \; \; a \mathbb Z + \frac{-b + \sqrt D}{2} \mathbb Z $$
Lehman's notation is different. Also, for real fields and indefinite forms the mapping is usually two to one; it's a long story.
The first form is the group identity, the second is a group generator.
$$ \langle 1, 43, 512 \rangle $$
$$ \langle 2, 43, 256 \rangle $$
$$ \langle 4, 43, 128 \rangle $$
$$ \langle 8, 43, 64 \rangle $$
$$ \langle 16, 43, 32 \rangle $$
$$ \langle 32, 43, 16 \rangle $$
$$ \langle 64, 43, 8 \rangle $$
$$ \langle 128, 43, 4 \rangle $$
$$ \langle 256, 43, 2 \rangle $$
Indeed, Dirichlet's method gives
$$ \langle 2, 43, 2^8 \rangle \circ \langle 2^k, 43, 2^{9-k} \rangle = \langle 2^{k+1}, 43, 2^{8-k} \rangle $$
There is a recent book by Lehman, where the entire book takes binary forms and quadratic fields side by side, illustrating the isomorphism over and over, as a lead-in to later studies in algebraic number theory
199: < 1, 1, 50> Square 199: < 1, 1, 50>
199: < 2, -1, 25> Square 199: < 4, 3, 13>
199: < 2, 1, 25> Square 199: < 4, -3, 13>
199: < 4, -3, 13> Square 199: < 5, 1, 10>
199: < 4, 3, 13> Square 199: < 5, -1, 10>
199: < 5, -1, 10> Square 199: < 2, 1, 25>
199: < 5, 1, 10> Square 199: < 2, -1, 25>
199: < 7, -5, 8> Square 199: < 7, 5, 8>
199: < 7, 5, 8> Square 199: < 7, -5, 8>
Best Answer
I have not checked your calculations and will just assume them to be correct.
By using the Minkowski bound, you can deduce that any ideal class contains an integral ideal of norm smaller than $9$.
Now recall that the norm is multiplicative and any integral ideal can be written uniquely as the product of prime ideals. Additionally $[P]$ is of order $9$ in the ideal class group, so by Lagrange's theorem the cardinality of the Class group is divisible by $9$. Now you can show that there are less than $18$ different ideals of small enough norm that you can generate as products of the prime ideals of norm less than $9$.
Thus the class number is $9$. Since it contains elements of order $9$ the class group is cyclic and hence isomorphic to $\mathbb{Z}/9\mathbb{Z}$.