The reason that a field must be used in Shamir's reconstruction scheme is
that the calculations used in the reconstruction need to divide one "number"
by another, and division is not defined in $\mathbb Z$, the set of integers:
$\frac{m}{n}$ is not necessarily a member of $\mathbb Z$.
So, why not use $\mathbb R$, or $\mathbb Q$ which can be "implemented" in
terms of pairs of integers? The answer again is that computers use
floating-point arithmetic which is not the same as real arithmetic, or
integer arithmetic which is, if we ignore overflow and underflow, effectively
modular arithmetic in $\mathbb Z_{2^m}$ which is not a field but a ring.
A more subtle issue is that the Shamir's scheme implicitly assumes that
a polynomial of degree $n$ with coefficients in a field does not
have more than $n$ roots in the field, which property is not true in rings.
For example, the polynomial $x^2 - 1$ has four roots $\pm 1, \pm 4$
in the ring $\mathbb Z_{15}$ instead of the two $\pm 1$
that it has in a field such as $\mathbb Z_{17} = \mathbb F_{17}$.
As a concrete example of what might happen with integer arithmetic as
implemented on a general-purpose computer, consider
this formula for secret reconstruction
$$s_0 = (-1)^k (x_1x_2x_3\cdots x_k) \sum_{i=1}^k \frac{y_i}{x_i\cdot c_i}$$
taken from another answer of mine.
Here, $s_0$ is the secret that is reconstructed from shares $(x_i,y_i)$
(that is, $y_i = s(x_i)$) and
$$c_i = (x_i-x_1)(x_i-x_2)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_k).$$
Now consider the case where the $k$ shareholders who have
gathered to reconstruct the secret all happen to have $x_i$ an odd integer.
Then, $c_i$ is an even integer -- in fact, a multiple of $2^{k-1}$ --
and so $\frac{y_i}{x_i\cdot c_i}$ is not necessarily an integer. However, the sum
$s_0$ will work out to be an integer.
With ordinary integer arithmetic on a computer, the fractional parts
of $\frac{y_i}{x_i\cdot c_i}$, if any, will
be lost when the integer division indicated is computed, and thus
$s_0$ will not be computed correctly. This is not to say that one could
not manage this issue
with careful programming that works around the problem, but we also have
to deal with the possibility that the computations might cause overflow
or underflow which also needs to be worked around. In any case, there
can be problems that arise because the polynomials re-constructed
via Lagrange interpolation are not necessarily the same as the ones used to
construct the secret originally. Foe example, both $x^2-1$ and
$(x-1)(x-4) = x^2-5x+4$ have roots $1$ and $4$ in $Z_{15}$. Since
we don't know ahead of time which shares will be available for
reconstruction, we cannot be sure whether we will reconstruct the
correct polynomial in the Lagrange interpolation process.
Thus, whether the secret recovery process will work as claimed
in a ring is an open question. That the process will work in
a field is guaranteed.
