Not following what's happening with the exponents in this proof by mathematical induction. I'm not understanding what's happening in this proof. I understand induction, but not why $2^{k+1}=2*2^{k}$, and how that then equals $k^{2}+k^{2}$. Actually, I really don't follow any of the induction step - what's going on with that algebra? Thanks!
Let $P(n)$ be the statement
$2^n > n^2$, if $n$ is an integer greater than 4.
Basis step:
$~~~~~P(5)$ is true because $2^5 = 32 > 5^2 = 25$
Inductive step:
$~~~~~~$Assume that $P(k)$ is true, i.e. $2^k > k^2$. 
We have to prove that $P(k+1)$ is true. Now
$$\begin{aligned}
2^{k+1} & = 2 \cdot 2^k\\
& > 2 \cdot k^2\\
& = k^2 + k^2\\
& > k^2 + 4k\\
& \geq k^2 +2k+1\\
& = (k+1)^2
\end{aligned}$$
$~~~~~$ because $k>4$
$~~~~~$ Therfore $P(k+1)$ is true
Hence from the principle of mathematical induction the given statement is true. 

You ask two questions:
(1) By definition, $$2^{k+1} = \underbrace{2 \quad \textrm{...} \quad 2}_{\textrm{ (k+1 times)}} = 2 \cdot \underbrace{2 \quad \textrm{...} \quad 2}_{\textrm{ (k times)}} = 2 \cdot 2^k.$$ As others have pointed, it's the usual rule to calculate an exponential: $a^x \cdot a^y = a^{x+y}$.
(2) The step immediately before $k^2 + k^2$ is that it equal $2k^2$. I assume that is clear (?) - you obviously always have $2a = a+a$. These two are equal; they are not equal to $2 \cdot 2^k$ (as you write, but probably don't mean). So I assume you ask why is the step before true, the one going $2 \cdot 2^k > 2 \cdot k^2$. This is using the induction assumption, assumed to be true for $k$ (as you write yourself earlier in "Let P(n) be the statement..."): if $2^k > k^2$, then obviously also $2 \cdot 2^k > 2 \cdot k^2$
