A bit more detailed look at what Qiaochu said. Unfortunately my answer won't really be expressed in terms of $r$ and $s$. I hope it still helps you in some way.
We know that $\Bbb{F}_p[X]\otimes \Bbb{F}_{p^n}\cong\Bbb{F}_{p^n}[X]$ and that $\Bbb{F}_{p^n}$ is a flat $\Bbb{F}_p$-module. Let us consider the short exact sequence
$$0\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]/\langle r\rangle\to0,$$
where the first map is multiplication by $r$, and the last module is isomorphic to $\Bbb{F}_{p^m}$. Upon tensoring with $\Bbb{F}_{p^n}$ this gives rise to the short exact sequence
$$0\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]/\langle r\rangle\to0.$$
Therefore a comparison of the last modules shows that
$$
\Bbb{F}_{p^m}\otimes \Bbb{F}_{p^n}\cong \Bbb{F}_{p^n}[X]/\langle r\rangle.
$$
The polynomial $r$ has no multiple zeros in $\overline{\Bbb{F}_p}$, so over $\Bbb{F}_{p^n}$ it factors into a product of distinct factors
$$
r=\prod_{i=1}^t r_i
$$
for some irreducible polynomials $r_i\in\Bbb{F}_{p^n}[X]$. Because these factors are distinct, the Chinese remainder theorem tells us that
$$
\Bbb{F}_{p^n}[X]/\langle r\rangle\cong\bigoplus_i \Bbb{F}_{p^n}[X]/\langle r_i\rangle.
$$
Note that everything above applies equally well to any finite extension of fields $L/K$. There is no need for the fields $L,K$ to be finite. We only needed the polynomial $r$ to be separable, so that we avoided the possibility of repeated factors.
The next step is, as Qiaochu pointed out, specific to Galois extensions. Namely, we can also deduce that the factors $r_i$ are Galois conjugates of each other. Most notably they all have the same degree. In the case of finite fields we can see this more concretely, because we know that the Galois group consists of powers of the Frobenius automorphism $F:x\mapsto x^p$.
The zeros of $r$ are
$$
\alpha,\alpha^p,\alpha^{p^2},\ldots,\alpha^{p^{m-1}}
$$
where $\alpha$ is some (fixed) zero of $r$. For example $\alpha=X+\langle r\rangle$. The roots of one of the factors $r_i$ are then lists like
$$
\alpha^{p^i},\alpha^{p^{i+n}},\alpha^{p^{i+2n}},\ldots
$$
because we get such lists of conjugates by applying powers of $F^n$ to one of them.
The original list of $m$ roots consisted of a single orbit of the Galois group $G=\langle F\rangle$. This list is now partitioned into orbits of the
subgroup $H=\langle F^n\rangle$. Basic facts about actions of cyclic groups tells us that the $H$-orbits all have size $m/\gcd(m,n)$, and that there are
$\gcd(m,n)$ of them. Therefore we get
$$
\Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}\cong\bigoplus_{i\in D}\Bbb{F}_{p^n}(\alpha^{p^i}),
$$
where the set $D=\{0,1,\ldots,\gcd(m,n)-1\}$ consists of representatives of those orbits. It is easy to see that all those fields
$$\Bbb{F}_{p^n}(\alpha^{p^i})\cong \Bbb{F}_{p^\ell}$$
with $\ell=\operatorname{lcm}(m,n)$.
Summary: $\Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}$ is isomorphic to a direct sum of $\gcd(m,n)$ copies of $\Bbb{F}_{p^\ell}$ where $\ell=\operatorname{lcm}(m,n)$. In particular, $\Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}$ is a field if and only if $\gcd(m,n)=1$.
This actually brings up a subtle point. What do we mean by $5$ in a finite field? Or if you choose to define $5$ in terms of $1 ~(5=1+1+1+1+1)$, then what do we mean by $1$?
One answer is to define $5$ in terms of equivalence classes. Say that two integers $m$ and $n$ are equivalent if $p \vert (m-n).$ First, you prove this really is an equivalence relation on the integers. Then you define $[m]+[n]=[m+n]$ and $[m][n]= [mn]$. So by $5$ we actually mean the equivalence class $[5]$.
You need to prove that your field operations are well-defined (you get the same answer no matter which representative of an equivalence class you choose) and that $[0]$ and $[1]$ really are the additive and multiplicative identities, as you'd expect. But once you've done that, you can see that $[4]+[1]=[5]$ (and usually we abuse notation by dropping the brackets) because we've defined it that way.
Best Answer
There aren't any, since the left side is $0$ at $x=1$.
But one can get a general description of $f(x)$ and $g(x)$ if you replace the right-hand side by, say, $x-1$ (or equivalently $x+2$).