Let $Q = (Q_1, \dots, Q_n)$ be a random Gaussian vector, where $Q_i$'s are i.i.d. from $N(0, \sigma^2)$. Consider a sequence $\{Q^k\}_{k=1}^{K}$, and suppose that $Q^k$'s are independent random Gaussian vectors.
What can be said about the following probability? $$\mathbb{P}\left[\left\vert \sum_{k=1}^{K-1} \langle Q^k, Q^{k+1} \rangle \right\vert \ge t\right]$$
By triangle inequality and the fact that for any reals $a_1,\ldots,a_{K-1}$, $\left[\sum_{k=1}^{K-1} a_k \ge t\right]\implies \bigcup_{k=1}^{K-1}\left[ a_k \ge t/(K-1)\right] $ we get $$ \mathbb{P}\left[\left\vert \sum_{k=1}^{K-1} \langle Q^k, Q^{k+1} \rangle \right\vert \ge t\right] \le \sum_{k=1}^{K-1} \mathbb{P}\left[\left\vert \langle Q^k, Q^{k+1} \rangle \right\vert \ge t\right] = (K-1)\mathbb{P}\left[\left\vert \langle Q^1, Q^{2} \rangle \right\vert \ge t/(K-1)\right] $$ Where we used the union bound and the fact that the inner products have the same distribution. Now rename the pair $(Q^1,Q^2)$ as $(X,Y)$ for convenience and note that by polarization identity $$\frac{2}{\sigma^2}\langle X,Y\rangle = (\|X+Y\|^2_2 - \|X-Y\|^2_2)/2\sigma^2 $$ Where $\|\cdot\|_2$ is the standard Euclidean norm. But now since $X$ and $Y$ are independent $\mathcal N(0,\sigma^2\mathbf I_n)$ variables, we have that $(X+Y)/\sqrt 2\sigma $ and $(X-Y)/\sqrt 2\sigma $ are independent $\mathcal N(0,I_n)$ variables (you can easily check that their covariance is $0$).
This tells us that $\frac{2}{\sigma^2}\langle X,Y\rangle = Z - Z'$ where $Z$ and $Z'$ are i.i.d. $\chi^2$ random variables with $n$ degrees of freedom, hence we know that it follows a symmetric about $0$ variance-gamma distribution with parameters $\lambda = n/2$, $\alpha =1/2$ and $\beta=\mu=0$.
Back to our original problem, we get that for all $t'>0$ $$\begin{align}\mathbb{P}\left[2\left\vert \langle X, Y \rangle \right\vert/\sigma^2 \ge 2t'/\sigma^2\right]&\le \mathbb{P}\left[2 \langle X, Y \rangle/\sigma^2 \ge 2t'/\sigma^2\right] +\mathbb{P}\left[2 \langle X, Y \rangle/\sigma^2 \le -2t'/\sigma^2\right]\\ &= 2\mathbb{P}\left[2 \langle X, Y \rangle/\sigma^2 \ge 2t'/\sigma^2\right]\\ &= 2 (1-F_\gamma\left(2t'/\sigma^2\right))\end{align} $$
Where $F_\gamma$ is the CDF of the variance-gamma distribution. Unfortunately, no closed-form expression is known for this CDF, but section 2.2 of this survey provides some bounds and asymptotics.
Letting $t'\equiv t/(K-1)$ we finally find $$\mathbb{P}\left[\left\vert \sum_{k=1}^{K-1} \langle Q^k, Q^{k+1} \rangle \right\vert \ge t\right] \le 2(K-1)\cdot\left(1-F_\gamma\left(\frac{2t}{(K-1)\sigma^2}\right)\right) $$
Another way to look at this is to observe that $\langle X,Y\rangle = \sum_{i=1}^n X_i Y_i$ is sub-exponential as a sum of $n$ products of (sub)-Gaussian random variables. Therefore a bound similar to the one I gave in my answer to your previous question is also possible.
