With three points, you can either do a linear interpolation which is $\mathcal{O}(h)$ or a quadratic interpolation which is $\mathcal{O}(h^2)$.
The linear interpolation is simple. Let $\theta \in [0,1]$ (the percentage along the interval from $x$ to $x+h$). Then,
$$ f(x+\theta h) = \theta f(x) + (1-\theta)f(x+h) + \mathcal{O}(h)$$
It's just the weighted average between the two points, putting more weight on the end you're closer to. Notice that this only uses two points. Also notice that if $\theta = \frac{1}{2}$, we recover the midpoint formula you had before.
The quadratic interpolation is exactly how it sounds: you interpolate by making a polynomial. Recall that in the general form we have that a quadratic is
$$f(x) = ax^2 + bx + c$$
Thus we have 3 unknowns and 3 equations: $f(x) = y_1$, $f(x+h) = y_2$, and $f(x+k) = y_3$. Given that $k$, $h$, and the $y_i$ are known, this uniquely defines the quadratic through these points, and thus gives the coefficients $a$, $b$, and $c$. In this interval, we can then define $\theta \in [0,1] $ (the percentage to $k$) and approximate via
$$ f(x+\theta k) = a (x + \theta k)^2 + b (x + \theta k) + c + \mathcal{O}(k^2) $$
Notice the order here is 2, so this will be a slightly better approximation than the linear interpolation.
Let's do the math out for completeness. We get the equations:
$$ y_1 = a x^2 + bx + c $$
$$ y_2 = a x^2 + (b + 2ah)x + (c + bh + a h^2) $$
$$ y_3 = a x^2 + (b + 2ak)x + (c + bk + a k^2) $$
The easier way to do this is write everything in terms of the coefficent we wish to find:
$$ y_1 = x^2 a + x b + c $$
$$ y_2 = (x^2 + 2hx + h^2) a + (x + h) b + c $$
$$ y_3 = (x^2 + 2kx + k^2) a + (x + k) b + c $$
So this is a linear system:
$$ \left[\begin{array}{c}
y_{1}\\
y_{2}\\
y_{3}
\end{array}\right]=\left[\begin{array}{ccc}
x^{2} & x & 1\\
x^{2}+2hx+h^{2} & x+h & 1\\
x^{2}+2kx+k^{2} & x+k & 1
\end{array}\right]\left[\begin{array}{c}
a\\
b\\
c
\end{array}\right] $$
The solution is then
$$ \left[\begin{array}{c}
a\\
b\\
c
\end{array}\right]=\left[\begin{array}{ccc}
x^{2} & x & 1\\
x^{2}+2hx+h^{2} & x+h & 1\\
x^{2}+2kx+k^{2} & x+k & 1
\end{array}\right]^{-1}\left[\begin{array}{c}
y_{1}\\
y_{2}\\
y_{3}
\end{array}\right] $$
I threw this into Mathematica to get:
$$ \left[\begin{array}{c}
a\\
b\\
c
\end{array}\right]= \left(
\begin{array}{c}
\frac{y_1 (h-k)-h y_3+k y_2}{h k (h-k)} \\
\frac{y_1 (-(h-k)) (h+k+2 x)+h y_3 (h+2 x)-k y_2 (k+2 x)}{h k (h-k)} \\
\frac{(k+x) \left(y_1 (h-k) (h+x)+k x y_2\right)-h x y_3 (h+x)}{h k (h-k)} \\
\end{array}
\right)
$$
as the explicit equations for the coefficients.