Following the idea of your derivation: let $t_i$ be $1$ if a run of heads starts at coin number $i$ ($i=1 \cdots n$), $0$ otherwise. Then $X_n = \sum_{i=1}^n t_i$ and
$$
P(t_i=1)=E(t_i)=
\begin{cases}
p & \text{if $i=1$} \\
p(1-p) & \text{otherwise} \\
\end{cases}
$$
The vars $t_i$ are not independent, but $E(X_n)=\sum E(t_i) = p + (n-1)p (1-p)$ as you already found. Now let's compute
$$E(X^2_n)= E\left[\left(\sum t_i \right)^2\right]=\sum_{i=1}^n\sum_{j=1}^n E(t_i t_j)$$
The terms of the sum can be grouped in:
A) $i=j$: (with $n$ terms) $E(t_i t_j)=E(t_i^2)=E(t_i)$. The sum over these terms is $p + (n-1)p (1-p)$.
B) $|i-j|=1$: (with $2 \times (n-1)$ terms). Here one of the two must be zero,so $E(t_i t_j)=0$
In the remainder, the variables $t_i,t_j$ are independent. Hence
C) $|i-j|>1$ and $i=1$ or $j=1$ : (with $2\times(n-2)$ terms) $E(t_i t_j) = p^2(1-p)$
D) $|i-j|>1$ and $i>1$ and $j>1$ : (with $n^2-5n +6$ terms) $E(t_i t_j) = p^2(1-p)^2$
Finally
$$E(X_n^2)=p+\left( n-1\right) \,\left( p-1\right) \,p+ 2\,\left( n-2\right) \,\left( 1-p\right) \,{p}^{2}+\left( {n}^{2}-5\,n+6\right) \,{\left( 1-p\right) }^{2}\,{p}^{2}$$
$$Var(X_n) = E(X_n^2) - E(X_n)^2 = \left( 1-p\right) \,p\,\left( 3\,n\,{p}^{2}-5\,{p}^{2}- 3 \, n\,p+3 p+n\right)=\\
=\left( 1-p\right) \,p\,\left[ n\,\left( 3\,{p}^{2}-3\,p+1\right) -p\,\left( 5\,p-3\right) \right]$$
As pointed out by Did in a comment, this is valid for $n\ge 4$; if $n<4$ the group D vanishes; if $n\le 2$, only case A survives. Then
$$Var(X_1)=p(1-p) \hspace{9pt} Var(X_2)=p(1-p)^2(2-p)\hspace{9pt} Var(X_3)=p(1-p)^2(4p^2-6p+3)$$
I ask you for additional explanation. Meanwhile I'll post here another approach.
Denote by $\tau_i^5$ the random variable that counts the time required to get five heads starting from $i$ heads, ok?
What we want is exactly $E[\tau_0^5]$, right?
Now, you can evaluate $E[\tau_0^5]$ conditioning at the first step.
$$
E[\tau_0^5] = \frac{E[\tau_0^5]}{2} + \frac{E[\tau_1^5]}{2} +1
$$
Explaining the equation above. With probability $1/2$ you have a tail, so the time you will take to get five heads is the same, because you have any heads. On the other hand, with the same probability you get a head, and now, the number of flips needed to get 5 heads is $E[\tau_1^5]$, because now you that you have one head. The +1 it is because you have to count the first step.
Repeating the argument above you get
$$
E[\tau_1^5] = \frac{E[\tau_0^5]}{2} + \frac{E[\tau_2^5]}{2} +1
$$
Proceeding this way, and remembering $E[\tau_5^5]=0$, you get
$$
E[\tau_0^5] = 62
$$
This may seems more complicated at the first sight, but the idea of "to conditionate at what happens at the first time (or movement)" solve a big variety of problems.
Best Answer
I'd use indicator random variables. For $1\leq j\leq n-1$, let $Z_j$ take the value $1$ if the $j$th and $j+1$st coin tosses are different, and take the value $0$ otherwise. Then the number of runs is $R=1+\sum_{j=1}^{n-1}Z_j$ and its expected value is $$\mathbb{E}(R)=1+ \sum_{j=1}^{n-1}\, \mathbb{E}(Z_j)=1+(n-1) 2p(1-p).$$ I will leave the calculation $\mathbb{E}(Z_j)=2p(1-p)$ as an exercise!