As has already been pointed out, it might be worthwhile learning a bit of algebraic geometry first but I'll try to answer this without going too deep. For most of this I will follow Silverman, as I think elliptic curves make divisors a lot more approachable than abitrary varieties or schemes.
Recall some basics of topology, every compact orientable 2-manifold looks like (is homeomorphic to) an $n$-torus, i.e. either a sphere or a doughnut shape with $n$ holes. This $n$ happens to be equal to the genus of this surface. How do we connect this to elliptic curves though?
Over $\mathbb{C}$, the solution set of an elliptic curve actually looks like a torus topologically (this is called the uniformisation theorem), as we quotient $\mathbb{C}$ by some lattice. Since this is simply a torus with 1 hole, the genus of an elliptic curve is $1$. We can also do this for hyperelliptic curves (although the genus will be different).
On to divisors:
Every place of the funtion field of an elliptic curve corresponds to a point on this curve, so we can just think of a divisor as a formal sum of points. The tiny catch is that this sum needs to be finite but that is perfectly fine.
For example, if $E: y^2=x^3+1$ over $\mathbb{C}$, then $Q=(2,3)$ and $R=(0,1)$ are points on $E$. We then have divisors
$D_1 = 2[R], D_2=-2[R]+[Q], D_3=[R]+[Q]$, and these form an abelian group under the obvious addition so $D_1+D_2=D_3$. (Forget about the group law on elliptic curves if you know about it). We also have the concept of the degree of a divisor, which is just the sum of the $n_i$. In the above the degrees of $D_1,D_2$ and $D_3$ are $2,-1$ and $1$ respectively.
Now consider a nonzero function $f$ in the function field of the curve. We say $v_P(f)=n$ if $f$ has a zero of order $n$ at $P$ (note $n$ can be negative if it has a pole instead. Checking for all poles and zeroes of this function, we can construct a divisor out of $f$, which is called $div(f)$.
For example, let $f=x-2$. Then $f=0$ if $x=2$ which occurs for the points $Q=(2,3)$ and $Q'=(2,-3)$ and we can easily see that these are simple zeroes. This means $v_Q(f)=v_{Q'}(f)=1$.
We can also see that $f$ does not have a pole at any point in the affine plane, however, elliptic curves really live in projective space and it turns out that $f$ has a pole of order $2$ at $\infty$. So $v_{\infty}(f)=-2$.
Having found all the zeroes and poles, we now see that the divisor of $f$ is $div(f)=[Q]+[Q']-2[\infty]$. Note that the degree of this is zero and this is always the case for divisors of functions.
For the Riemann Roch space $\mathcal{L}(D)$, we want to find all functions whose divisor is at least the negative of another divisor. For example, if we take $D=[Q]+[Q']$, then we are saying that we are only allowing functions which have at most simple poles at $Q$ and $Q'$ and nowhere else. It is an easy check that this forms a complex vector space.
Adjusting our idea above, $\dfrac{1}{x-2}$ lives in this space (as well as any scale multiple by the vector space structure), but no others do. Hence, we have $1$ basis element so $l(D) \geq 1$. We can also check that $\dfrac{x}{x-2}$ lives in here and that is it so $l(D)=2$. It is useful to note that since all nonconstant functions have poles, then if the degree of $D$ is negative then $l(D)=0$.
Now the Riemann Roch theorem can either be viewed as the definition for the genus, or if you already know this, it can be used to find $l(D)$, up to some correction term.
Now $\mathcal{K}$ is what is known as the canonical divisor and comes from a differential form which I won't attempt to explain here, but has degree $2g-2$.
So in the above, $\mathcal{K}-D$ has degree $-2$ which is negative hence $l(\mathcal{K}-D)=0$ and we can see that the Riemann Roch formula holds.
Best Answer
Let $D$ be a divisor of degree 6 and let $P$ be an arbitrary point on $X$. Then Riemann-Roch applied to $D$ and $D-P$ gives the following equations: $$l(D)-l(K-D)=2,$$ $$l(D-P)-l(K-D+P)=1.$$ Subtracting, we get $$l(D)-l(D-P)+l(K-D+P)-l(K-D)=1,$$ and now we have some observations to make:
Let's look more closely at $K-D+P$ and $K-D$. These are divisors of degree $2g-2-6+1=3$ and $2g-2-6=2$, respectively, and so we have $l(K-D+P)\leq 3$ and $l(K-D)\leq 2$ (ref). But $l(K-D)\neq 2$, as otherwise that's a $g_2^1$, so we must have $l(K-D)\leq 1$ and therefore also $l(K-D+P)\leq 2$. But again, $l(K-D+P)\neq 2$, as that would be a $g_3^1$, and $X$ doesn't have one of those. So $l(K-D+P)\leq 1$ and $l(K-D)\leq 1$.
Therefore if we can find a $D$ so that $l(K-D)=1$, we will have that $l(K-D+P)=1$ as well, and this choice of $D$ is base-point free. That can be done without too much trouble: take an effective divisor $P_1+\cdots+P_8$ linearly equivalent to $K$, then take $D=P_1+\cdots+P_6$.