The question has been asked here. I know the method so hints in the comments/answers don't help. The problem is that the process stops from a point on! And it must not be so because the Descent Procedure guarantees otherwise...
$34 \cdot 1973 = 259^2 + 1^2; \\ (21^2 + 1^2)(259^2 + 1^2)=(21 \cdot 259 +1)^2+(259-21)^2 \implies 13 \cdot 1973 = 7^2 + 160^2. $ It's good because by the method must $13 \le 34/2$ as it is.
$(7^2 + 4^2)(7^2 + 160^2)=(7 \cdot 7 +160 \cdot 4)^2+(160 \cdot 7-4 \cdot 7)^2 \implies 5 \cdot 1973 = 53^2 + 84^2. $ Good again since $5 \le 13/2$.
$(3^2 + 4^2)(53^2 + 84^2)=(99 \cdot 5 )^2+(8 \cdot 5)^2 \implies 5 \cdot 1973 = 99^2 + 8^2.$ And $5 \le 5/2$ doesn't hold. Also from $5 \cdot 1973 = 99^2 + 8^2 \implies 5 \cdot 1973 = 53^2 + 84^2$ by the method! That famous circle of misfortune here!
Two problems here:
$1.$ It must always $m \le M/2$ as we go from $M \cdot p = A^2 + B^2$ to $m \cdot p = a^2 + b^2$. Why it can't be true in this case though it contradicts an existing theorem about it?
$2.$ Then how we arrive at $1973 = 23^2 + 38^2$ starting from $34 \cdot 1973 = 259^2 + 1^2$?
Added - Same problem occurs when trying to arrive at $96493 = 258^2 + 173^2$ starting from $10 \cdot 96493 = 261^2 + 947^2$!
If I understood the question correctly, you want a procedure to write $1973$ as a sum of two squares using the facts that $$ \begin{aligned} 2\cdot17\cdot1973&=259^2+1^2,\\ 17&=4^2+1^2,\\ 2&=1^2+1^2. \end{aligned} $$
I would also approach this using complex numbers, Gaussian integers $\Bbb{Z}[i]=\{a+bi, a,b\in\Bbb{Z}\}$, to be more precise. The idea is to use the multiplicativity of the norm function $N(a+bi)=a^2+b^2$. To a great extent the method depends on the fact that $N$ is a Euclidean norm of $\Bbb{Z}[i]$ (If we dig a bit deeper it turns out to be a chicken-egg question whether the theory of sums of two squares was first or the fact that $\Bbb{Z}[i]$ has unique factorization).
Anyway, $2=(1+i)(1-i)=-i(1+i)^2$ is a factor of $(259+i)(259-i)$, so we expect $259+i$ to be divisibile by $1+i$. A basic complex division shows that this is true: $$ \frac{259+i}{1+i}=\frac{(259+i)(1-i)}{(1+i)(1-i)}=\frac{260-258i}2=130-129i. $$ That went smoothly, we now know that $$ 17\cdot1973=130^2+129^2. $$ Let's get rid of that $17$. It is the product $17=(4+i)(4-i)$. Those factors are primes of $\Bbb{Z}[i]$ because their norms are primes of $\Bbb{Z}$. Again we expect $130-129i$ to be divisible by one of them. Let's check $$ \frac{130-129i}{4-i}=\frac{(130-129i)(4+i)}{(4-i)(4+i)}=\frac{649}{17}-\frac{386}{17}i. $$ This didn't work. So we need to try the other prime of norm $17$: $$ \frac{130-129i}{4+i}=\frac{(130-129i)(4-i)}{(4+i)(4-i)}=\frac{391-646i}{17}=23-38i. $$
That's better!