Question: I have the following two matrices:

1) $$\begin{bmatrix}1&0&0\\ 0&1&1\\ 0&0&0\\0&0&0 \end{bmatrix}$$

I believe this matrix is in the form of reduced row echelon form but I'm a bit unsure.

2) $$\begin{bmatrix}1&2&6&0&0\\0&0&1&2&-2\\0&0&0&0&1\end{bmatrix}$$

I believe this matrix is in row echelon form.

I'm still not 100% sure what defines a row echelon form and reduced row echelon form. I've tried looking everywhere for a clear understanding of it, but I just could grasp it 100%.

Answer: 
The matrices you have are in row echelon form for they satisfy the following conditions:

1. All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix).
2. The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be $1$.).
3. All entries in a column below a leading entry are zeroes (implied by the first two criteria.).

However, the second matrix is not in reduced row echelon form because on top of being in row echelon form it also needs to satisfy:

* Every leading coefficient is 1 and is the only nonzero entry in its column.

So, as mentioned by NasuSama, the second can be simplified to reduced row echelon form.
The first matrix is already in reduced row echelon form. The second matrix can be simplified to

$$\begin{bmatrix}
1 & 2 & 0 & -12 & 0\\
0 & 0 & 1 & 2 & 0\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}$$