The largest $a$ is $4/9$.

We first show that $a = 4/9$ is admissible. For each $2 \le k \le n$ by the Cauchy-Schwarz Inequality, we have
$$(x_{k-1} + (x_k - x_{k-1})) \left( \frac{(k-1)^2}{x_{k-1}} + \frac{3^2}{x_k - x_{k-1}} \right)^2 \ge (k-1+3)^2,$$
which can be rewritten as
$$\frac{9}{x_k - x_{k-1}} \ge \frac{(k+2)^2}{x_k} - \frac{(k-1)^2}{x_{k-1}}. \quad (1)$$
Summing Eq. (1) over $k = 2, 3, \dots, n$ and adding $9/x_1$ to both sides, we have
$$9 \sum_{k=1}^{n} \frac{1}{x_k - x_{k-1}} \ge 4 \sum_{k=1}^{n} \frac{k+1}{x_k} + \frac{n^2}{x_n} > 4 \sum_{k=1}^{n} \frac{k+1}{x_k}.$$
This shows the original inequality holds for $a = 4/9$.

Next, we show that $a = 4/9$ is the optimal choice. Consider the sequence defined by $x_0 = 0$ and $x_k = x_{k-1} + k(k+1)$ for $k \ge 1$, that is,
$$x_k = \frac{1}{3}k(k+1)(k+2).$$
Then the left-hand side of the original inequality equals
$$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1},$$

$$\begin{align*} a \sum_{k=1}^{n} \frac{k+1}{x_k} &= 3a \sum_{k=1}^{n} \frac{1}{k(k+2)} \\ &= \frac{3}{2}a \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+2} \right) \\ &= \frac{3}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) a. \end{align*}$$
$$$$
When $n$ tends to infinity, the left-side tends to $1$ while the right-hand side tends to $\frac{9}{4}a$. Therefore $a$ has to be at most $4/9$.

Hence the largest value of $a$ is $4/9$.