Given the Cauchy's problem: $y'' = 1, y(0) = 0, y'(0) = 0$. Why finite difference method doesn't agree with recurrence equation? The domain we want to obtain the solution on is  $x \in [0,1]$.
Let's write the second difference equation corresponding to this differential equation problem $\frac{y_{i-1}-2y_i+y_{i+1}}{h^2} = 1$ (is that step valid by the way?). The first condition is $
y_0 = 0 $. For the second condition we use the first forward difference: $\frac{y_1-y_0}{h} = 0$, so $y_1 = y_0 = 0$. We choose $h = 0.1111$ for the grid. Then we solve $Ay = b$,
where the matrix $A = \begin{bmatrix}
1  & 0     & 0      &  0 &    \dots   &   0     \\
0  & 1 & 0 & 0 &  \dots  &         0 \\
1 & -2 & 1 & 0  &  \dots  &    0\\
 0  & 1 & -2 & 1 & \dots &  0  \\
\vdots  &        & \ddots & \ddots & \ddots   & \vdots \\
0 & \dots      &  0      & 1     &  -2 &      1
\end{bmatrix},$
$b = [0, 0, h^2, h^2, \dots,h^2]^T.$
This is a script (you could play with it), solving this problem:
k = 10;
x = linspace(0, 1, k);
h = (x(end) - x(1)) / (k - 1);
D = sparse(1 : k, 1 : k, 1 * ones(1, k), k, k);
Elow = sparse(2 : k, 1 : k-1, -2 *ones(1, k-1), k, k);
Elow2 = sparse(3 : k, 1 : k - 2, ones(1, k - 2), k, k);
S = Elow + D + Elow2;
S(1,1) = 1;
S(1,2) = 0;
S(2,1) = 0; S(2,2) = 1; S(2,3) = 0;
Sf = full(S);
b = h^2 * ones(1,k)'; b(1) = 0; b(2) = 0;
u = S \ b;

Increasing the number of $k$ we can see that finite difference solution looks like the analytical solution $y = 0.5 x^2$
On the other hand we can solve $y_{i-1}-2y_i+y_{i+1} = 1, y_0 = 0, y_1 =0$ directly as 2nd order nonhomogeneous recurrence equation, the solution would be $y_i = 0.5(i^2 - i)$.
ps I know that the RK4 method is OK for such problem, but I'm trying to study different numerical methods, so I took finite difference method for this equation to be sure that it's working.

Summing up, I have several questions:
is this the case when my finite difference scheme should converge to actual solution?
why the recurrence equation solution gives so bad result? Is it correct actually?

The problem with a direct recurrence is that you write the recurrence wrong. It is 
$$y_{i-1}-2y_i+y_{i+1} = h^2$$
therefore the solution is $y_i = 0.5(i^2 - i)h^2$. Now recall that $x = ih$, or $i = \frac{x}{h}$, so $y(x) = 0.5(x^2 - hx)$, and as $h$ tends to $0$, the approximation tends to the solution as expected.
