Suppose that we have the equation $x'=\left(\begin{array}{cc} 2 & 1 \\ 1 & 2\end{array}\right)x$. If you solve for the eigenvalues of this matrix, they are $1$ and $3$ with corresponding eigenvectors $v_{1}=\left(\begin{array}{c} 1 \\ -1 \end{array}\right)$ and $v_{2}=\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$ respectively. So the general solution will look like \[x(t)=c_{1}v_{1}e^{t}+c_{2}v_{2}e^{3t}.\] Note that since both eigenvalues are positive, so both vectors here are being multiplied by $e$ to the power of some positive number times $t$, hence $x(t)$ is moving away from the origin as $t\rightarrow \infty$ and converging toward the origin as $t\rightarrow-\infty$. Let's consider the trajectories of the solutions when $c_{1}=0,c_{2}\neq 0$ and $c_{2}=0,c_{1}\neq 0$.
If $c_{1}=0$, then all solutions will be moving parallel to the vector $v_{2}$, their direction depends on the sign of $c_{2}$: if $c_{2}>0$, then solutions move in the same direction as (i.e. parallel with) $v_{2}$; if $c_{2}<0$, then it's the opposite direction. We denote this in our picture by showing arrows pointing outward from the origin along the lines parallel to $v_{1}$ and $v_{2}$ respectively.
If $c_{1}=0$, this corresponds to $x(0)$ lying on the blue line (since $x(0)$ will be equal to a constant multiple of $v_{2}$). Then $x(t)$ will move along the blue line away from the origin (that is, in the direction of the arrows). Similarly, $c_{2}=0$ corresponds to $x(0)$ being contained in the green line, and its solution will move in the direction of the arrow from its starting position.
Now, suppose $x(0)$ is in the shaded quadrant as shown in the figure by the black dot. Then note that for large $t$, $e^{3t}$ will be much larger in comparison to $e^{t}$ because of the higher exponent, hence \[x'(t)=c_{1}v_{1}e^{t}+3c_{2}v_{2}e^{3t}\approx 3c_{2}v_{2}e^{3t}.\] Thus, as $t$ gets large, the direction that $x$ is moving in (i.e. $x'$) is roughly the same direction as $v_{2}$ (i.e. almost parallel with the blue line). Similarly, as $t\rightarrow -\infty$, \[x'(t)=c_{1}v_{1}e^{t}+3c_{2}v_{2}e^{3t} \approx c_{1}v_{1}e^{t},\] so that for small $t$, $x$ is moving toward the origin (as we explained earlier), but roughly in parallel with the vector $v_{1}$ (that is, almost parallel with the green line).
If we plotted a solution with $x(0)$ in each quadrant, the solutions would look like these:
The important thing to remember here is that for large $t$, the solutions are moving away from the origin and moving parallel to the vector $v_{2}$ (i.e. parallel to the blue line), and for small $t$, the solutions are moving into the origin and parallel to $v_{1}$ (i.e. parallel to the green line).
Now consider the derivative: \[x'(t)=-c_{1}u_{1}e^{-t}+3c_{2}u_{2}e^{3t}.\] Note that as $t\rightarrow\infty$, the second term is much larger than the first (the first goes to zero, in fact), so \[x'(t)\approx 3c_{2}u_{2}e^{3t}\] and if $t$ is a very negative number, $e^{3t}$ will be very small, and so \[x'(t)\approx -c_{1}u_{1}e^{-t}.\] Hence, for very negative values of $t$, $x(t)$ almost lies on the green line parallel to $u_{1}$ (since it is approximately a constant multiple of $u_{1}$) and moves in the direction $-c_{1}u_{1}$ (i.e. the opposite direction of $u_{1}$). Then for large values of $t$, $x$ is moving in the direction $u_{2}$ and lies very close asymptotically to the blue line parallel to $u_{2}$ (since it is approximately a constant multiple of $u_{2}$). Thus, the picture of some typical solutions should look something like this:
