In standard Kalman filter calculations. Namely, the prediction step is given by
$\hat{x}_{k|k-1} = F_k\,\hat{x}_{k-1|k-1} + B_k\,u_k, $
$P_{k|k-1} = F_k\,P_{k-1|k-1}\,F_k^\top + Q_k$
Under this setup,
$K_k = P_{k|k-1}\,H_k^\top S_k^{-1}, \\ \hat{x}_{k|k} = \hat{x}_{k|k-1} + K_k\,\tilde{y}_k, \\ P_{k|k} = (I - K_k\,H_k)\,P_{k|k-1}$
While this is fine, in my case the equation of Kalman filter includes contemporaneous variable. More specifically, below is my equation for Kalman filter
${x}_{k|k-1} = {x}_{k-1|k-1} + {y}_{k|k-1} + W_{1, k}, \\ {y}_{k|k-1} = {y}_{k-1|k-1} + W_{2, k} \\ W_{1, k} \sim N \left( 0, {\sigma}_{1} \right), W_{2, k} \sim N \left( 0,{\sigma}_{2} \right)$
My question is, under this setup do I need to make any adjustment in the expressions of $\hat{x}_{k|k}, P_{k|k}$ above?
Any pointer will be very helpful
In your question you were jumping ahead a little. Namely, you combined the dynamics with stochastic terms together with the prediction step, but the prediction step should be deterministic. And in one of your comments you mentioned that the actual considered dynamics is given by
\begin{align} x_{k} &= x_{k-1} + y_{k} + w_{1,k} \tag{1} \\ y_{k} &= y_{k-1} + w_{2,k} \tag{2} \end{align}
with $w_{1,k} \sim N\!\left(0, \sigma_1\right)$ and $w_{2,k} \sim N\!\left(0, \sigma_2\right)$.
When writing this into one state space equation it is important which state vector one chooses. Two logical options in my opinion would be either $z_k = \begin{bmatrix}x_k & y_k\end{bmatrix}^\top$ or $z_k = \begin{bmatrix}x_k & y_{k+1}\end{bmatrix}^\top$.
For the first option one has to substitute $(2)$ into $(1)$
$$ z_k = \begin{bmatrix}x_k \\ y_k\end{bmatrix} = \begin{bmatrix}x_{k-1} + y_{k-1} + w_{2,k} + w_{1,k} \\ y_{k-1} + w_{2,k}\end{bmatrix} = \underbrace{\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}}_{F_k} z_{k-1} + \underbrace{\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix}w_{1,k} \\ w_{2,k}\end{bmatrix}}_\text{noise}. \tag{3} $$
It can be noted that $(3)$ would be equivalent to replacing the noise term by just a noise vector but with covariance matrix
$$ Q_k = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \sigma_1^2 + \sigma_2^2 & \sigma_2^2 \\ \sigma_2^2 & \sigma_2^2 \end{bmatrix}. $$
For the second option one just have to update the indices of $(2)$
$$ z_k = \begin{bmatrix}x_k \\ y_{k+1}\end{bmatrix} = \begin{bmatrix}x_{k-1} + y_{k} + w_{1,k} \\ y_{k} + w_{2,k+1}\end{bmatrix} = \underbrace{\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}}_{F_k} z_{k-1} + \underbrace{\begin{bmatrix}w_{1,k} \\ w_{2,k+1}\end{bmatrix}}_\text{noise}. \tag{3} $$
Since the noise terms are stationary processes the covariance matrix of it would simply be
$$ Q_k = \begin{bmatrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \end{bmatrix}. $$
If it is also known what is measured (and it is linear in the state) and what noise is added to it, then a normal Kalman filter can be used. Using either state space representation should yields equivalent results.