You can think of $L$ as the set of all things of the form $ar^2+br+c$ where $a,b,c$ come from ${\bf F}_2$ and $r$ satisfies $r^3+r+1=0$. Now $L$ also contains a zero of $X^3+X^2+1$, so you are looking for the values of $a,b,c$ such that if $s=ar^2+br+c$ then $s^3+s^2+1=0$. You can just multiply everything out, use $r^3+r+1=0$ to get it down to a quadratic in $r$, set the coefficients to zero, and solve. This is probably a mess. There may be an easier way to do it, but this will get you an element of $L$ that satisfies $X^3+X^2+1=0$. Once you have that element, you know the isomorphism you're looking for takes $r$ to that element. Since $r$ generates $L$, you get the entire isomorphism.
Now in fact I think you have managed to find that (in my notation) $r+1$ is the element you are looking for. So your map from $L$ to $L'$ takes $r+1$ in $L$ to a generator, call it $t$, in $L'$. Might be easier to see it as a map from $L'$ to $L$, taking $at^2+bt+c$ to $a(r+1)^2+b(r+1)+c$.
A (base-$g$) discrete logarithm of a finite field $\Bbb{F}_q$, is a function
$$
\log_g:\Bbb{F}_q^*\to\Bbb{Z}_{q-1}
$$
defined via the equivalence $g^j=x\Leftrightarrow \log_g(x)=j$. For this to be well-defined it is imperative that $g$ is a primitive element, i.e. a generator of $\Bbb{F}_q^*$, and that the domain of $\log_g$ is the
residue class ring of integer modulo $q-1$, as $g^{q-1}=g^0=1$.
It immediately follows that the discrete logarithm satisfies the familiar rules
$$
\begin{aligned}
\log_g(x\cdot y)&=\log_g(x)+\log_g(y),\\
\log_g(x^n)&=n\cdot\log_g(x)
\end{aligned}
$$
for all elements $x,y\in \Bbb{F}_q^*$ and all integers $n$. The arithmetic
on the r.h.s. is that of the ring $\Bbb{Z}_{q-1}$.
It is known that when $q=8$, a zero $\alpha$ of $x^3+x+1$ generates $\Bbb{F}_8^*$. This is proven by the following calculation, where we repeatedly
use the fact that we are working in characteristic two, and that we have the
relation $\alpha^3=\alpha+1$.
$$
\eqalign{
\alpha^0&=&&=1,\\
\alpha^1&=&&=\alpha,\\
\alpha^2&=&&=\alpha^2,\\
\alpha^3&=&&=1+\alpha,\\
\alpha^4&=&\alpha\cdot\alpha^3=\alpha(1+\alpha)&=\alpha+\alpha^2,\\
\alpha^5&=&\alpha\cdot\alpha^4=\alpha(\alpha+\alpha^2)=\alpha^2+\alpha^3=\alpha^2+(1+\alpha)&=1+\alpha+\alpha^2,\\
\alpha^6&=&\alpha\cdot\alpha^5=\alpha(1+\alpha+\alpha^2)=\alpha+\alpha^2+\alpha^3=
\alpha+\alpha^2+(1+\alpha)&=1+\alpha^2,\\
\alpha^7&=&\alpha\cdot\alpha^6=\alpha(1+\alpha^2)=\alpha+\alpha^3=\alpha+(1+\alpha)&=1.
}$$
We see from the end results in the last column that all the non-zero
quadratic polynomials evaluated at $\alpha$ appear. This is yet another confirmation of the fact that $\alpha$ is a primitive element.
The discrete logarithm is used to replace the cumbersome multiplication (and raising
to an integer power) of the field with more familiar integer arithmetic. Exactly like the old-timers used logarithm tables to replace the error-prone multiplication with the easier addition.
For example
$$
(1+\alpha)(1+\alpha+\alpha^2)=\alpha^3\cdot\alpha^5=\alpha^8=\alpha^7\cdot\alpha=\alpha.
$$
Note that both the base-$\alpha$ discrete logarithms and its inverse mapping are needed. I generate such a table as a part of the program initialization, whenever I carry out extensive computer-aided calculations involving a finite field. The above table gives the discrete logarithm when read from right to left, and the inverse mapping (that we actually produced above) when read from left to right.
Similarly with $q=16$ we use $\gamma$, a zero of $x^4+x+1$. This time the table looks like
$$
\begin{aligned}
\gamma^0&=&1\\
\gamma^1&=&\gamma\\
\gamma^2&=&\gamma^2\\
\gamma^3&=&\gamma^3\\
\gamma^4&=&\gamma+1\\
\gamma^5&=\gamma(\gamma+1)=&\gamma^2+\gamma\\
\gamma^6&=\gamma(\gamma^2+\gamma)=&\gamma^3+\gamma^2\\
\gamma^7&=\gamma^4+\gamma^3=&\gamma^3+\gamma+1\\
\gamma^8&=(\gamma^4)^2=&\gamma^2+1\\
\gamma^9&=\gamma(\gamma^2+1)=&\gamma^3+\gamma\\
\gamma^{10}&=\gamma^4+\gamma^2=&\gamma^2+\gamma+1\\
\gamma^{11}&=&\gamma^3+\gamma^2+\gamma\\
\gamma^{12}&=\gamma^4+\gamma^3+\gamma^2=&\gamma^3+\gamma^2+\gamma+1\\
\gamma^{13}&=\gamma^4+\gamma^3+\gamma^2+\gamma=&\gamma^3+\gamma^2+1\\
\gamma^{14}&=\gamma^4+\gamma^3+\gamma=&\gamma^3+1\\
(\gamma^{15}&=\gamma^4+\gamma=&1)
\end{aligned}
$$
Thus for example
$$
(\gamma^3+1)(\gamma^2+1)=\gamma^{14}\cdot\gamma^8=\gamma^{22}=\gamma^7=\gamma^3+\gamma+1.
$$
As another example of the use of this table I want to discuss the problem of factorizing $x^4+x+1$ over $\Bbb{F}_4$. To that end we need to first identify a copy of $\Bbb{F}_4$ as a subfield of $\Bbb{F}_{16}$. We just saw that $\gamma$ is of order fifteen. Therefore $\gamma^5=\gamma^2+\gamma$ and $\gamma^{10}=\gamma^2+\gamma+1$ are third roots of unity. It is then trivial to check that we have a homomorphism of fields $\sigma:\Bbb{F}_4\to\Bbb{F}_{16}$ given by $\sigma(\beta)=\gamma^5$. Note that composing this (from either end) by the Frobenius automorphism gives an alternative embedding $\beta\mapsto \gamma^{10}$.
Basic Galois theory tells us that
$$
x^4+x+1=(x-\gamma)(x-\gamma^2)(x-\gamma^4)(x-\gamma^8)
$$
as we get the other roots by repeatedly applying the Frobenius automorphism $F:x\mapsto x^2$. Here we see that the factor
$$
(x-\gamma)(x-\gamma^4)=x^2+x(\gamma+\gamma^4)+\gamma^5=x^2+x+\gamma^5
$$
is stable under the automorphism $F^2$, and thus (as we also see directly!) has its
coefficients in the subfield $\sigma(\Bbb{F}_4)$. The same holds for the remaining factor
$$
(x-\gamma^2)(x-\gamma^8)=x^2+x(\gamma^2+\gamma^8)+\gamma^{10}=x^2+x+\gamma^{10}.
$$
Pulling back the effect of $\sigma$ we get the desired factorization
$$
x^4+x+1=(x^2+x+\beta)(x^2+x+\beta+1)
$$
in $\Bbb{F}_4[x]$.
Here is a local version of similar tables for $\Bbb{F}_{256}$
Best Answer
There are two ways you might want to represent a field of order $q^k$ (where $q\in \mathbb{N}$ is a prime and $k \gt 0$ a positive integer).
One is to imagine a simple algebraic field extension of $\mathbb{Z}_q$ which has a basis $\{1,\alpha,\ldots,\alpha^{k-1}\}$, namely $\mathbb{Z}_q[\alpha]$ where $\alpha$ satisfies an irreducible monic polynomial of degree $k$ over $\mathbb{Z}_q$:
$$ \alpha^k + c_1 \alpha^{k-1} + \ldots + c_{k-1}\alpha + c_k = 0 $$
Any element of this field extension can be represented in terms of the basis, and the addition is just the same as the addition of the $k$-dimensional vector space over $\mathbb{Z}_q$. But when we multiply, we have to perform substitutions to eliminate powers $\alpha^k$ and higher by using:
$$ \alpha^k = - c_1 \alpha^{k-1} - \ldots - c_{k-1}\alpha - c_k $$
The other way to think about it is to start with an irreducible polynomial $p(x)$ of degree $k$ over $\mathbb{Z}_q$ and construct $\mathbb{Z}_q[x]/p(x)$ as a quotient ring. Since the ideal generated by $p(x)$ is maximal, the quotient ring is a field and has dimension $k$ over $\mathbb{Z}_q$ as a vector space.
These two constructions are equivalent, with $\alpha$ being a root of:
$$ p(x) = x^k + c_1 x^{k-1} + \ldots + c_{k-1}x + c_k $$
and identified with $x \bmod{p(x)}$ in $\mathbb{Z}_q[x]/p(x)$.
It turns out that all field extensions of degree $k$ over $\mathbb{Z}_q$ are isomorphic, so as a computational convenience the irreducible polynomial $p(x)$ may be chosen to be simple in some way (e.g. having as few nonzero coefficients as possible).
It's not hard to come up with a monic irreducible polynomial of degree $3$ over $\mathbb{Z}_2$ in order to construct a field of $8$ elements. You only need to check for divisibility by linear (first degree) factors to be sure of irreducibility.