If the field is countable, then I claim that there is such a
family of size continuum, which is the largest it could possibly
be, since $K^{\mathbb{N}}$ has size continuum.

To see this, we shall build a tree $T\subset K^{\lt\mathbb{N}}$ in
the space of finite sequences from $K$ so as to ensure the
following properties:

 - the tree is splitting, in the sense that every node has
 incomparable extensions.
 - any two distinct paths through $T$ have at most finite agreement.
 - any finitely many paths through $T$ are linearly independent.

If we can do this, then the collection $X$ of all paths through $T$
will be a size-continuum linearly independent family with any two
elements having at most finite agreement.

We build the tree recursively, by specifying it up to increasingly high finite levels. We
will ensure that distinct paths have at most finite agreement by
ensuring that on any given level of the tree, distinct nodes use
different elements of $K$ on that level. So once two paths through
the tree depart, the elements of $K$ that appear on them will be
totally different.

We can ensure that the tree is splitting by periodically inserting
a splitting level into the tree, where we make sure that every
node on that level has two successors in the tree.

Finally, the difficult part, we ensure the linear independence property. The key is to
realize that there are only countably many possible linear
dependence equations. So at a given stage of constructing the
tree, we may consider just one equation, which has the form, say,
$k_0f_0+\cdots+k_n f_n=0$, where each $k_i\in K$ is nonzero. The
point now is that for any $n$ distinct branches through the finite
tree as constructed so far, up to some height $N$, we may extend
the tree a bit more so that the extension of those branches those
branches no longer satisfy this equation in the finite dimensional vector space of sequences of that length. In finitely many steps,
I can take care of all the finitely many ways of having chosen $n$
distinct branches in the tree up to the previously constructed
level, and thereby handle them all. So at the $n^{th}$ stage of
the construction, I kill off this equation as a possible
dependence for the branches through my tree $T$. In the end, all
such dependencies are killed, and any finitely many branches
through the tree are linearly independent, completing the
argument.

A set-theoretic way to think about the argument is this: if you
force to add mutually generic sequences, then they will be
linearly independent. The construction of the tree amounts to a
density argument in forcing, and any sufficiently generic
collection of branches will satisfy the independence property. The tree argument is reminiscent of the fact that in the forcing extension obtained by adding a single Cohen real $V[c]$, there is a family of continuum many finitely-mutually-generic Cohen reals, which I explained in my answer to [Reals added after Cohen forcing](http://mathoverflow.net/a/99025/1946). 

When $K$ is uncountable, then I believe that set-theoretic issues will likely arise. For example, probably under Martin's Axiom you can still get large families for fields of size less than the continuum, using MA to choose the family of functions by using the fact that each dependency equation corresponds to a dense set in the forcing I have in effect described.