I suspect there is no closed form, but here's how you can calculate this for small values of $k$. I will be talking about binary strings, but it's the same problem.
If $S_k$ is the set of winning binary $k$-strings, consider the generalized question of, given a finite set $S$, deciding whether any given string contains a string from $S$. This is the problem of recognizing a regular language, and is solved by constructing a finite-state automaton corresponding to the set $S$. (See any reference on FSAs and the problem of matching regular expressions, etc.)
Following an idea I learnt from "Analytic Combinatorics" by Flajolet and Sedgewick, it is very easy to construct a generating function for a regular language (a language described by an FSA). Let $L_\alpha$ be the generating function corresponding to the state $\alpha$, so that the $n$-th coefficient counts the number of length-$n$ input strings such that the FSA, started from state $\alpha$, will finish in a terminal state. Let $\bar Q$ be the set of terminal states, and let $q(\alpha, a)$ be the transition function that gives the next state, where $\alpha$ is the current state and $a$ is the next input letter (belonging to the FSA's alphabet). Then the generating functions satisfy
$$ L_\alpha(z) = [\alpha\in\bar Q] + z\sum_{\text{letter }a} L_{q(\alpha, a)}(z). $$
If we denote by $T$ the adjacency matrix of the directed graph of transitions of the FSA ($T_{\alpha\beta} = \sum_{a}[q(\alpha,a)=\beta]$), we get
$$ \mathbf{L}(z) = v + z T \mathbf{L}(z), $$
where $\mathbf{L}=(L_\alpha)_\alpha$ and $v_\alpha=[\alpha\in\bar Q]$. This is a system of linear equations in $L$, which can be solved for $L_0$, the generating function for the initial state. Then $L_0$ is the generating function for the number of strings of length $n$ that belong to the regular language recognized by the FSA.
A short implementation (hopefully correct) leads me to the following generating functions for small $k$. Here $g_k(z)$ is the generating function for the set of binary strings that contain a winning $k$-string.
$$ g_2(z) = \frac{1}{1-2z}+\frac{1+z}{-1+z+z^2}. $$
$$ g_3(z) = -z^2+\frac{1}{1-2z}+\frac{1+z+z^2}{-1+z+z^3}. $$
$$ g_4(z) = -z^3+\frac{1}{1-2z}+\frac{-1-z-z^2-z^3+z^4+z^5}{1-z-z^2-z^4+z^6}. $$
$$ g_5(z) = -z^3-5 z^4+\frac{1}{1-2z}+\frac{-1-z-2 z^2-2 z^3-2 z^4+z^5+z^6+2 z^7+z^8+z^9}{1-z-z^3-2 z^5+z^8+z^{10}}. $$
$$ g_6(z) = \frac{1}{8 (-1+z)}-z^4-6 z^5+\frac{1}{8 (1+z)}+\frac{1}{1-2z}+\frac{\left(4+3 z+9 z^2+11 z^3+22 z^4+19 z^5+8 z^6+6 z^7+6 z^8-2 z^9-9 z^{10}-10 z^{11}-z^{13}+3 z^{15}+4 z^{16}+3 z^{17}\right)}{\left(4 \left(-1+z+z^3+2 z^5+3 z^6+2 z^7+z^9+z^{11}-3 z^{12}+z^{18}\right)\right)}. $$
$$ g_7(z) = -z^4-6 z^5-22 z^6+\frac{1}{1-2z}+\left(\begin{aligned}1&+z+2 z^2+3 z^3+5 z^4+6 z^5+5 z^6-4 z^7-5 z^8-9 z^9-10 z^{10}\\&-17 z^{11}-17 z^{12}-14 z^{13}-4 z^{14}+4 z^{15}+8 z^{16}+11 z^{17}+19 z^{18}\\&+13 z^{19}+12 z^{20}+2 z^{21}-4 z^{22}-5 z^{23}-8 z^{24}-9 z^{25}-5 z^{26}\\&-5 z^{27}+z^{29}+z^{30}+2 z^{31}+2 z^{32}+z^{33}+z^{34}\end{aligned}\right)/\left(\begin{aligned}-1&+z+z^3+z^5+z^6+5 z^7+z^8-3 z^{10}-z^{11}-4 z^{12}-z^{13}-10 z^{14}\\&-5 z^{15}-z^{16}+3 z^{17}-z^{18}+6 z^{19}+10 z^{21}+3 z^{22}\\&-z^{24}-4 z^{26}-5 z^{28}+z^{33}+z^{35}\end{aligned}\right)$$
For $n\geq k$, $k=3,4,5,6$, the coefficients of $g_k-\frac{1}{1-2z}$ reproduce A000930, A118647, A120118 and A133551.
Also, $g_2$ matches the answer by Jean-Sebastien (A008466).
Edit. Here is also $g_8$ (I realize it's a mess):
$$ g_8(z) = -z^5-7 z^6-29 z^7+\frac{1}{1-2 z} + \frac{\sum_{j\geq0}p_{8,j}z^j}{\sum_{j\geq0}q_{8,j}z^j}, $$
$$ \begin{aligned}(p_{8,j})_{0\leq j\leq 69} = \{&-1,-1,-1,-2,-3,-5,-7,-2,19,22, 16,15,8,17,40,10,-42,-76,\\&-73,-39,-12,-24,-52,-18,54,141,100,33,-16,-1,37,35,-41,\\&-146,-56,-19,14,14,-24,-38,45,104,12,14,-18,-21,14,13,-36,\\&-49,1,-3,18,17,-4,0,15,15,0,-1,-8,-7,0,0,-2,-2,0,0,1,1\}.\end{aligned} $$
$$ \begin{aligned}(q_{8,j})_{0\leq j\leq 70} = \{&1,-1,-1,0,-1,0,1,-3,-8,0,8, 8,10,5,-8,2,28,15,-25,-24,-28,\\&-24,19,18,-51,-40,55,16,55,45,-51,-36,61,45,-70,16,-67,\\&-40,70,19,-56,-24,58,-24,56,15,-56,2,28,5,-28,8,-28,0,28,\\&-3,-8,0,8,0,8,-1,-8,0,1,0,-1,0,-1,0,1\}.\end{aligned} $$
Here $g_8-\frac1{1-2z}$ matches A212398.
