To answer this question, we'll first want to step back and discuss what a finite field is, and what a representation is.
A finite field is a group of finite size (always $p^k$, for some prime $p$ and integer $k$, in this case, $2^8$), along with addition and multiplication operators, where certain identities are true (e.g. $a \times (b + c) = a \times b + a \times c$), see this link for the list.
It turns out that, given a specific size $p^k$, the field is essentially unique (in the sense that, if we have two different sets of the same size where they both preserve the identities, they will always be a mapping between the two sets which preserve both the addition and multiplication operations, hence those two sets are essentially the same.
This is nice, however we need a way to talk about these elements; what we do is assign the element names; for $GF(2^8)$, we have precisely 256 elements, and so we typically give each element a unique 8 bit name. Such a way of assigning names is known as the 'representation'.
For $GF(2^8)$, there are a number of reasonable ways to assign these names, one such way is to select an irreducible polynomial, and compute the multiplication modulo that polynomial. That is the approach that AES takes, with the irreducible polynomial $x^8 + x^4 + x^3 + x + 1$. Let us call this specific representation (which is a way of assigning names to field members) $A$.
However, that isn't the only way of assigning names; we could consider a different irreducible polynomial, say, $x^8 + x^4 + x^3 + x^2 + 1$. That gives a different way of assigning names to field elements, that is, a different representation, we may call this representation $B$.
Now, we define $L$ to be the function that takes a field element (using the $A$ naming convention), and return the name of that same field element (using the $B$ naming convention). It turns out $L$ isn't unique; we'll ignore that for now. It turns out that $L$ is an isomorphism, as the field it maps from (the $A$ representation) and the field it maps to (the $B$ representation) are the same (because they're both different ways of given labels to the same underlying field).
It turns out that, if we just look at $L$ as a mapping between 8 bits to 8 bits, it is bit-wise linear.
In the $A, B$ representations I just gave, one such mapping is:
$$L(01) = 01$$
$$L(02) = 03$$
$$L(04) = 05$$
$$L(08) = 0F$$
$$L(10) = 11$$
$$L(20) = 33$$
$$L(40) = 55$$
$$L(80) = FF$$
Because $L$ is linear, it is easy to deduce $L(X)$ for any other $X$ not listed.
And, what $+_A$ means is the addition operation in the $A$ representation; that is, $X +_A Y$ takes the two field elements $X, Y$ (named using the $A$ convention), and evaluates to the field addition of those elements (also expressed in the $A$ naming convention. Hence, when we say:
$$L(X+_AY)=L(X)+_BL(Y)$$
$X, Y$ are both arbitrary field elements (in the $A$ naming convention), and what we're saying is that the $L$ mapping preserves addition; that is, if we add two arbitrary elements $X, Y$ (in the $A$ representation) and then map the sum, we always get the exact same value as we would have if we first mapped $X$ and $Y$ into the $B$ representation, and then added them (in the $B$ representation).
We call out $+_A$ and $+_B$ because they aren't necessarily the same operation (actually, in this case they are, but not in general); $+_A$ assumes the inputs are in the $A$ representation, and $+_B$ assumes they are in the $B$ representation.
You asked for a numeric example, here's a simple one (using the $\times$ operator; it turns out that the addition operation isn't that interesting).
If we pick $X = 03$ and $Y = 05$, we have (on the left side):
$L(03 \times_A 05) = L(0F) = L(01) \oplus L(02) \oplus L(04) \oplus L(08) = 08$
(Note: we have $L(0F) = L(01) \oplus L(02) \oplus L(04) \oplus L(08)$ because of the bitwise linearity of $L$)
On the right side, we have:
$L(03) \times_B L(05) = (L(01) \oplus L(02)) \times_B (L(01) \oplus L(04)) = 02 \times_B 04 = 08$
That is, both sides come up with the same value. This will always be true no matter what values we picked for $X, Y$
