This is just weighted least squares and here is how I would approach it. To keep my notation simple I'll just have polynomials of order 1. It's trivial to extend the approach to polynomials of any order. Let $a_x$, $b_x$ and $a_y$, $b_y$ be the 'true' coefficients describing the path. So the path is
$$ (x(t), y(t)) = ( a_x + b_x t, a_y + b_y t). $$
For convenience I'll pack the coefficients into a column vector
$$p_0 = \left[\begin{array}{cccc} a_x & b_x & a_y & b_y \end{array}\right]^\prime $$
where $'$ denotes transpose.

The path is observed at a number of points in time, $t_1, \dots,t_n$ say, and these observations as subject to noise with known covariance matrices $P_1,\dots,P_n$. That is, the observations are pairs
$$(x_i, y_i) = (x(t_i), y(t_i)) + W_i$$
where $W_1,\dots,W_n$ are bivariate random variables $W_i = (X_i, Y_i)$ with individual covariances given by the $2\times 2$ matrices $P_1, \dots, P_n$. I'll assume that the $W_1, \dots, W_n$ are independent.

We can write the observations in vector form as
$$f = T p_0 + w$$
where
$$ f = \left[ \begin{array}{c} x_1 \newline y_1 \newline x_2 \newline y_2 \newline \vdots \newline x_n \newline y_n \end{array}\right]^\prime \qquad
T = \left[ \begin{array}{cccc}
1 & t_1 & 0 & 0 \newline
0 & 0 & 1 & t_1 \newline
1 & t_2 & 0 & 0 \newline
0 & 0 & 1 & t_2 \newline
\vdots & \vdots & \vdots & \vdots \newline
1 & t_n & 0 & 0 \newline
0 & 0 & 1 & t_n
\end{array} \right] \qquad
w = \left[ \begin{array}{c} X_1 \newline Y_1 \newline X_2 \newline Y_2 \newline \vdots \newline X_n \newline Y_n \end{array}\right]^\prime.$$
Let $P$ be the $2n\times 2n$ covariance of $w$. So $P$ is block diagonal with diagonals given by the $2\times 2$ matrices $P_1,\dots,P_n$.

You take a weighted least squares approach to estimation, that is, your estimators are given by the minimisers of the quadratic form
$$(f - Tp)^\prime P^{-1} (f - Tp). $$
The minimiser is given by
$$ \begin{array}{ll}
\hat{p} &= (T^\prime D T)^{-1} T^\prime D f \newline
&= M (Tp_0 + w) \newline
&= p_0 + Mw \newline
\end{array}$$
where $D = P^{-1}$ and $M = (T^\prime D T)^{-1} T^\prime D$. So the error in your coefficients is given by $\hat{p} - p_0 = Mw$ and the covariance of the error is
$$ C = \operatorname{cov}(Mw) = M \operatorname{cov}(w) M^\prime = (T^\prime D T)^{-1}. $$

You want to know the covariance of the error at time $t$, that is you want the covariance of
$$\left[\begin{array}{c} \hat{a}_x + \hat{b}_x t \newline \hat{a}_y + \hat{b}_y t \end{array}\right] - \left[\begin{array}{c} a_x + b_x t \newline a_y + b_y t \end{array}\right] \qquad \text{where} \qquad \hat{p} = \left[\begin{array}{cccc} \hat{a}_x & \hat{b}_x & \hat{a}_y & \hat{b}_y \end{array}\right]^\prime. $$
This is given by the covariance of
$$ \left[\begin{array}{cccc} 1 & t & 0 & 0 \newline
0 & 0 & 1 & t \end{array}\right] (\hat{p} - p_0) = K(t)(\hat{p} - p_0) = K M w $$
say. This is $K(t) (T^\prime D T)^{-1} K(t)^\prime$.