For every integer $k$ coprime to 1722, how can one compute the sum
$\sigma(k)= \sum_{1\leq b\leq 1722, (1,b)=1722} \lfloor\frac{kb}{1722}\rfloor (-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor}$?
(Here the sum is taken over integers $b$ between $1$ and 1722 coprime to 1722.)
An approach would be to consider a linear function $\tau(k)$ and compute $\sigma(k)-\tau(k)$ where
\begin{align*}\tau(k) &= \sum_{1\leq b\leq 1722, (1,b)=1722} \frac{kb}{1722}(-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor}\\ &=\frac{k}{1722} \sum_{1\leq b\leq 1722, (1,b)=1722}b\cdot (-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor} = -\frac{80\cdot 1724k}{1722} \end{align*}
It is not hard to see that $\sigma(k)-\tau(k)$ is bounded. I used Python to compute $\tau(k)$.
I would like to know how to compute both $\sigma(k)$ and $\tau(k)$ by hand.
You can use $\tau(k)$ to approximate $\sigma(k)$.
Your calculation of $\tau(k)$ was not correct. This is the correct function: $$\frac{k}{1722} \sum _b b (-1)^{-\left\lfloor \frac{b}{2}\right\rfloor -\left\lfloor \frac{b}{3}\right\rfloor -\left\lfloor \frac{b}{7}\right\rfloor -\left\lfloor \frac{b}{41}\right\rfloor +b}=-\frac{68960}{861}k$$
And this is a table to show how good is the approximation $$ \begin{array}{r|r|r} k & \sigma(k) & \tau(k)\\ \hline 100 & -8008 & -8009.29 \\ 200 & -16016 & -16018.6 \\ 300 & -24032 & -24027.9 \\ 400 & -32036 & -32037.2 \\ 500 & -40044 & -40046.5 \\ 600 & -48060 & -48055.7 \\ 700 & -56052 & -56065. \\ 800 & -64076 & -64074.3 \\ 900 & -72088 & -72083.6 \\ 1000 & -80092 & -80092.9 \\ 2000 & -160184 & -160186. \\ 3000 & -240276 & -240279. \\ 4000 & -320372 & -320372. \\ 5000 & -400464 & -400465. \\ 6000 & -480548 & -480557. \\ 7000 & -560648 & -560650. \\ 8000 & -640744 & -640743. \\ 9000 & -720832 & -720836. \\ 10000 & -800928 & -800929. \\ \end{array} $$