The largest possible $M$ for which the given property holds is $M = 2$.

We first show that the given property holds for $M = 2$. To do this, we show by induction on $n$ the stronger statement that $x_{n+1} > 2x_n > x_n + x_{n-1} + \dots + x_0$ for all $n \ge 0$.

For $n = 0$, this is the statement $x_1 > 2x_0 > x_0$ which translates with the given initial values to $3 > 2 > 1$.

Now suppose for the induction hypothesis that $x_{n+1} > 2x_n > x_n + x_{n-1} + \dots + x_0$. Then we find for $x_{n+2}$:
$$\begin{aligned} x_{n+2} &\ge 3x_{n+1} - (x_n + \dots + x_0) \\ &> 2x_{n+1} \\ &> x_{n+1} + x_n + \dots + x_0. \end{aligned}$$
This completes the induction step. By induction, it follows that for all sequences $x$ satisfying (a) and (b), the inequality $\frac{x_{n+1}}{x_n} > 2$ holds for all $n \ge 0$.

To show that we cannot find a higher value for $M$, we look at the sequence $x$ with $x_0 = 1$, $x_1 = 3$, and for which equality holds in (b), i.e. $x_0 + x_1 + \cdots + x_{n-1} = 3x_n - x_{n+1}$ for all $n \ge 1$. Then the following relation holds:
$$\begin{align*} x_{n+1} &= 3x_n - (x_{n-1} + \cdots + x_0) \\ &= 3x_n - x_{n-1} - (x_{n-2} + \cdots + x_0) \\ &= 3x_n - x_{n-1} - (3x_{n-1} - x_n) \\ &= 4x_n - 4x_{n-1}. \end{align*}$$
Note that this is a homogeneous linear recurrence relation, the characteristic equation of which is $\lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 = 0$. Since the characteristic equation has a double root at $\lambda = 2$, the general solution of the recurrence relation is of the form $x_n = B2^n + Cn2^n$ for real numbers $B$ and $C$.

If we now solve this for the given starting values $x_0 = 1$ and $x_1 = 3$, we get the system of equations $B + 0 = x_0 = 1$ and $2B + 2C = x_1 = 3$. Its unique solution is given by $B = 1$ and $C = \frac{1}{2}$. So the solution for these starting values is $x_n = 1 \cdot 2^n + \frac{1}{2} n 2^n = (n+2)2^{n-1}$.

Now that we have solved the recurrence relation, a simple computation yields
$$\frac{x_{n+1}}{x_n} = \frac{(n+3)2^n}{(n+2)2^{n-1}} = 2\frac{n+3}{n+2} = 2\left(1 + \frac{1}{n+2}\right).$$
So for sufficiently large $n$, this fraction becomes arbitrarily close to 2. To state this more precisely: suppose $M = 2 + \varepsilon$ with $\varepsilon > 0$. Then, for this sequence and $n > \frac{2}{\varepsilon} - 2$, we have $\frac{x_{n+1}}{x_n} = 2 + \frac{2}{n+2} < 2 + \varepsilon = M$. Therefore no such $M$ can have the given property.

So the largest value of $M$ that has the given property is 2, and in the first part of this solution we have already seen that $M = 2$ has the given property. $\square$