Call $f(n)$ the expected number of jumps needed to reach position $n$. If $n > 10$, then clearly
$$f(n) = 1 + \frac1{10}\sum_{d=1}^{10} f(n-d)\,,$$
since we make one jump of length $d \leq 10$ and then for the remaining distance we expect $f(n-d)$ jumps.
(Mathematically, this is Fubini's theorem for conditional expectations.)
For general $n \geq 1$, we have
$$f(n) = 1 + \frac1{\min(10, n)}\sum_{d=1}^{\min(10, n)} f(n-d)\,,$$
for the same reason, with the convention $f(0) = 0$. Thus, when $n \leq 10$ we have, by repeatedly substituting this recurrence relation,
$$f(n) = 1+\sum_{k=1}^{n} \sum_{n = n_1 > n_2 > \ldots > n_k \geq 1} \frac{n_k - 1}{n_1 \cdots n_k} \,.$$
(Maybe this can be simplified.) Explicitly,
$$\begin{align*}
f(1) &= 1 \\
f(2) &= 1 + \frac{1}2 = \frac32 \\
f(3) &= 1 + \frac13 \left( 1 + 1 + \frac12 \right) = \frac{11}6 \\
f(4) &= 1 + \frac14 \left( 1 + \frac32 + \frac{11}6 \right) = \frac{25}{12}
\end{align*}$$
etc. Setting $f(n) = g(n) + \frac{2n}{11}$, we obtain a homogeneous linear recurrence
$$g(n) = \frac1{10} \sum_{d=1}^{10}g(n-d) \qquad (n \geq 10)$$
with initial terms $g(0), \ldots, g(9)$, whose solution is of the form
$$g(n) = \sum_{i=1}^{10} c_i \lambda_i^n$$
where the $\lambda_i \in \mathbb C$ are the roots of the polynomial
$$10x^{10} - (x^9 + x^8 + \cdots + x + 1)$$
and the $c_i$ are determined by $g(0), \ldots, g(9)$ and can be computed in terms of the $\lambda_i$ by inverting a Vandermonde matrix.
Numerically, we see that all $\lambda_i$ have modulus at most $1$, so that
$$f(n) = \frac{2}{11} n + O(1) \,.$$
An explicit formula is only possible when we have an explicit formula for the $\lambda_i$, which is a notorious problem.
