How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion?
I think I know where your confusion comes from. Firstly, I would want to note an important point:

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Chemwiki.UCDavis

This is the characteristic that makes them different from ionic bonds. The surrounding atom will not become an ion in its full sense.
I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.
$\ce{H2O}$:
We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:

Source 1
Follow these instructions for &quot;normal&quot; molecules (the ones that abide octet in a perfect sense):

Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: $$6+(2*1)=8$$
I prefer to call this $\text{n}_v$. V represents valence shell.
Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get:
$$8+2*2=12$$
I prefer to call this $\text{n}_f$. F represents &quot;final&quot;. You can name it whatever you want.
If you want to know how many electrons are bonding and are shared between atoms, subtract $\text{n}_v$ from $\text{n}_f$:
$$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$
For an ease in calculation, I call this $\text{n}_b$. As you might have just figured out,
If you want to know how many es are nonbonding and &quot;belong&quot; to a single atom, subtract $\text{n}_b$ from $\text{n}_v$ to get:
$$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$
Call this whatever you want. I rather calling it $\text{n}_{nb}$.

With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:

Source 2
$\ce{NH4+}$
In ions, we have to take the charge of the ion into an account. Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from $\text{n}_v$.
So let's get on with this thing's lewis structure:

$\text{n}_v = 5+(4*1) -1=8$ (Notice (-1) that was because of a 1+ ion charge.)
$\text{n}_f = 8 + (4*2)=16$
$\text{n}_b = 16-8=8$
$\text{n}_{nb} = 8-8=0$

There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as dative covalent bonding. We get this structure in accordance to the calculations and thoughts:

Source 4
Finally, $\ce{NH2-}$
Now this case is easily solvable through the explanations for ammonium.

$\text{n}_v = 5 + 2 +1=8$ The +1 is because of the charge of the ion.
$\text{n}_f = 8 +2*2=12$
$\text{n}_b = 12-8=4$
$\text{n}_{nb} = 8-4=4$

Put some thinking, and you should get this (ignore the other ones if you like):

Source 5