The field is everywhere, the lines are a drawing

Put a small test charge $q_0$ at a point, measure the force on it, and divide:

$$\vec{E} = \frac{\vec{F}}{q_0}, \qquad \vec{F} = q\vec{E}, \qquad [\vec{E}] = \text{N/C}.$$

Three consequences follow immediately, and each one is a place students slip.

1. The field does not depend on the test charge. Double $q_0$ and the measured force doubles, so the ratio is unchanged. The field at a point is a property of the SOURCES, not of whatever you put there to detect it.
2. The test charge is positive by convention. So for a positive charge $\vec{F}$ is along $\vec{E}$, and for a negative charge $\vec{F}$ is opposite to $\vec{E}$. The sign appears exactly once, in $\vec{F} = q\vec{E}$.
3. A charge does not feel its own field. The field at a point due to a set of charges, used to compute the force on one of them, is the field of everything else.

For a single point charge the field is radial:

$$E = \frac{k|q|}{r^2}, \qquad \text{pointing away from } +q \text{ and toward } -q.$$

The field of several charges at a point is the vector sum of what each one contributes there:

$$\vec{E}_{\text{net}} = \sum_i \frac{kq_i}{r_i^2}\,\hat{r}_i.$$

The $\hat{r}_i$ is doing real work. Adding the magnitudes $kq_i/r_i^2$ as though they were numbers on a line is only valid when every contribution happens to point the same way. Three charges each giving $3000$ N/C at a point almost never give $9000$ N/C; they can give anything from $9000$ down to zero, and which one it is depends entirely on the angles.

The procedure that does not fail:

1. For each source, draw the arrow AT the field point: away from a positive source, toward a negative one.
2. Write each arrow's magnitude $k|q_i|/r_i^2$, using that source's own distance.
3. Resolve into $x$ and $y$ components with the geometry's angles.
4. Sum $x$ with $x$ and $y$ with $y$, then recombine: $E = \sqrt{E_x^2 + E_y^2}$.

Two configurations get memorized and then swapped, so keep them apart. At the midpoint between two EQUAL LIKE charges the two arrows point in opposite directions and cancel: $E = 0$. At the midpoint between $+q$ and $-q$ separated by $d$, the arrow from the positive charge points away from it and the arrow from the negative charge points toward it, which is the SAME direction, so they add:

$$E_{\text{mid}} = 2\cdot\frac{kq}{(d/2)^2} = \frac{8kq}{d^2}.$$

Field lines are drawn so that the tangent at each point gives the direction of $\vec{E}$ there, and the local density of lines encodes the magnitude: tightly packed means strong, spread out means weak. The rules that follow are worth stating.

1. Lines start on positive charge and end on negative charge, or run off to infinity.
2. Lines never cross. A crossing would give the field two directions at one point.
3. The number of lines leaving a charge is proportional to its magnitude, which is why a $+2q$ charge is drawn with twice the lines of a $+q$.
4. Lines meet a conductor's surface perpendicularly in electrostatic equilibrium.

The blank space between two drawn lines is not empty of field. The drawing samples a continuous vector field at finitely many places; a point that no line happens to pass through has a field you get by interpolating from its neighbors. Two points at the same distance from an isolated point charge have exactly the same field magnitude whether or not the artist put a line through either of them.

Zero field does show up in these diagrams, but as a place lines avoid symmetrically, such as the midpoint between two equal like charges, not as a gap between strokes of the pen.

Newton's second law connects the field to the ACCELERATION, not to the velocity:

$$\vec{a} = \frac{q\vec{E}}{m}.$$

A trajectory is fixed by the acceleration together with the initial velocity, so it coincides with a field line only in the special case of a charge released from rest in a field whose direction never changes along the motion. Fire a proton across a uniform field between two plates and it traces a parabola, exactly like a projectile fired horizontally in gravity: constant acceleration one way, constant velocity across it.

The curved case is the one to burn in. Release a charge from rest on a curved field line. It starts moving along the tangent, which is correct. One instant later it has velocity in that old direction while the field at its new position points somewhere slightly different, so its path bends away from the line and keeps diverging from it. Curved field lines are essentially never trajectories.

Say it as a rule: the field line tells you which way the charge is being pushed right now, and nothing about where it has been or where it will end up.