How does $p^2 = $ sum of squares $(p_x^2 + p_y^2 + p_z^2)$? Why does this happen algebraically? I'm sorry I am new to this, and I don't exactly understand the algebra.
Normally if $x = a + b$, then $x^2 = (a + b)^2 = a^2 + 2ab + b^2$, and not equal to $a^2 + b^2$. I don't quite understand why momentum, $p$, which can be split into $p_x$ and $p_y$ and $p_z$, becomes the sum of each individual part squared when $p$ is squared...
Moreover, as I understand, the "del" operator (upside down delta) is equal to the sum of partial derivatives with respect to $x$, $y$ and $z$ (let's call these $a + b + c$). However, del squared is not equal to $(a + b + c)^2$ but instead to $(a^2 + b^2 + c^2)$... How does this work? I just feel like I am doing random algebra so I would like to better understand, thank you.
Is it something to do with vectors and dot products..? Why do these become operators in quantum mechanics..?
 A: Let's start from scratch $x=a+b$ where a and b are both scalars. So $x^2 = (a+b)^2$
But momentum is a vectorial quantity defined as
$$\vec{P} = P_{x}\hat{i} + P_{y}\hat{j}+ P_{z}\hat{k}$$
and $P^2$ is nothing but $\vec{P}\vec{P}$, following products are zero
$$\hat{i}\cdot\hat{j} = 0 \hspace{4mm} , \hspace{4mm}\hat{i}\cdot\hat{k} =0 \hspace{4mm} , \hspace{4mm} \hat{j}\cdot\hat{k} = 0$$
and
$$\hat{i}\cdot\hat{i} = 1 \hspace{4mm} , \hspace{4mm}\hat{j}\cdot\hat{j} =1 \hspace{4mm} , \hspace{4mm} \hat{k}\cdot\hat{k} = 1$$
thus
$$P^2 = P_{x}^2 + P_{y}^2 + P_{z}^2$$
gradient of a function f is defined in the following manner
$$\nabla f= \Big(\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} ,\frac{\partial f}{\partial z}\Big) = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} +\frac{\partial f}{\partial z}\hat{k}$$
divergence of $\vec{v}$ vector field
$$\nabla\cdot \vec{v} =  \Big(\frac{\partial}{\partial x} , \frac{\partial}{\partial y} ,\frac{\partial}{\partial z}\Big)\cdot(v_{x} , v_{y} , v_{z}) = \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} +\frac{\partial v_{z}}{\partial z}$$
what you call as del squared is the Laplace operator and it's defined as divergence of a gradient (check the above divergence and gradient formulas and use the definition) thus Laplacian is written
$$\Delta = \nabla^2 = \nabla \cdot \nabla = \frac{\partial ^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2}$$
A: 
I don't quite understand why momentum, p, which can be split into p_x and p_y and p_z, becomes the sum of each individual part squared when p is squared...

Momentum $\vec p$ is not the sum of its  three components $p_x$, $p_y$ and $p_z$. Rather, it is a vector where each of these components points in the direction of the unit vector corresponding to each axis. That is
$$\vec p = p_x \hat i + p_y \hat j + p_z \hat j$$
This means if you take the dot product (all terms with differing unit vector products vanish)
$$\vec p . \vec p = {p_x}^2 + {p_y}^2 + {p_z}^2$$

However, del squared is not equal to (a + b + c)^2 but instead to (a^2 + b^2 + c^2)... How does this work?

Exactly the same reasoning above but with del
$$\nabla = \partial_x \hat i + \partial_y \hat j + \partial_z \hat k$$

Is it something to do with vectors and dot products..?
Why do these become operators in quantum mechanics..?

Yes as you can see from the explanation above.
And in quantum mechanics observables are replaced by (Hermitian) operators and these operators have derivatives in them. For example momentum in one dimension
$$p_x = -i \hbar \frac{\partial}{\partial x}$$
Why they have partial derivatives, I would recommend you read this.
A: In 3 dimensions the norm of a vector $\vec{u}$ is given by the equation $\left|\vec{u}\right|=\sqrt{{u_x}^2+{u_y}^2+{u_z}^2}$ with $\left|\vec{u}\right|$ being the norm of the vector and ${u_x}$,${u_y}$, ${u_z}$ being the components of the vector.  Momentum is a vector $\vec{p}=\left(p_x,p_y,p_z\right)$ so $\left|\vec{p}\right|=\sqrt{{p_x}^2+{p_y}^2+{p_z}^2}$ and $\sqrt{{p_x}^2+{p_y}^2+{p_z}^2}^2=\left(\left({p_x}^2+{p_y}^2+{p_z}^2\right)^\frac{1}{2}\right)^2={p_x}^2+{p_y}^2+{p_z}^2$.
A: The key element is that momentum is a vector, which has three components $\vec p = (p_x, p_y, p_z)$. There is nothing special for the momentum operator, it  behaves like any other vector. So, let's consider a vector $\vec p = (p_x,p_y)$ in 2D -- it's just simpler to draw.

Now, $p^2 =  \vec p \cdot \vec p$ is the projection of $\vec p$ onto itself. Hence, it's the length squared. In order to calculate the length of $\vec p$ we can either use Pythagoras theorem (because the two components are perpendicular to each another):
$$
|\vec p|^2 = p_x^2 + p_y^2
$$
or we could use the general formula
$$
\vec a \cdot \vec b = |a|^2 + |b|^2 + 2 \vec a \cdot \vec b
$$
with $\vec a = \vec p_x$ and $\vec b=\vec p_y$.  Since the two components are orthogonal, the cross term vanishes $2 \vec a \cdot \vec b = 0$.
As you see the derivative is not involved in the calculation of the length. The same calculation is true for any vector.
A: 
the components of the vector $~\vec p$ are:
$$\vec p=\begin{bmatrix}
  x \\
  y \\
  z \\
\end{bmatrix}$$
according to Pythagoras law :
$$p^2=c^2+z^2$$
with  $~c^2=x^2+y^2$
$\Rightarrow$
$$p^2=x^2+y^2+z^2$$
