To solve for the line you want, you really do the "basic" thing you might expect.
The formula for a line is $f(x) = ax + b$.
We know that $f(2) = 2a + b = 14$, and $f(4) = 4a + b = 25$.
It follows that $f(4) - f(2) = 2a = 11$, so $a = 11\times 2^{-1}\bmod 41$.
We have that $21\times 2\equiv 1\bmod 41$, so $ a = 11\times 21\bmod 41 = 26$.
You can then solve that $f(2) = 2\times 26 + b = 14\implies b \equiv 38$.
So $f(x) = 26x + 38$.
This argument works, but doesn't scale to higher-degree polynomials well.
To do this, it helps to know some linear algebra.
Specifically, the space of all polynomials of a given degree over a field $k$ $\mathcal{P}_n(k) = \{\sum_{i = 0}^n a_ix^i\mid a_i\in k\}$ forms a vector space over $k$.
This vector space is of dimension $n+1$.
One obvious basis for it is the monomial basis $\{1,x,\dots,x^{n+1}\}$.
For any set of $n+1$ points $x_1,\dots,x_{n+1}\in k$, there is a special matrix called the Vandermonde matrix.
It maps polynomials (written in the monomial basis) to the "evaluation basis", i.e. sends a polynomial $p(x) = \sum_{i=0}^n a_ix^i = (a_0,\dots,a_n)$ (in the monomial basis) to an "evaluation representation" $(p(x_0),\dots,p(x_{n+1}))$.
Anyway, the general way to reconstruct Shamir's secret sharing is
- collect enough evaluation points $(p(x_0),\dots,p(x_{n+1}))$
- use the (inverse) vandermonde matrix to change basis to the monomial basis $p(x) = \sum_{i=0}^n a_ix^i$
- evaluate $p(0)$.
There are optimizations you can do to this (you don't need to compute a full matrix-vector product, as you only need a single entry $p(0) = a_0$ from the result, so it suffices to compute an inner product), but the above is what I find most conceptually clear.
