Let's start with a formulation of probability states that is obviously adequate to the task, and then consider whittling their number down for a more efficient computation.
Let $s(L,T)$ denote a state where $L$ is a three character string recording the last three throws:
$$ L \in \{\; hhh,\; hht,\; hth,\; htt,\; thh,\; tht,\; tth,\; ttt\; \} $$
and $T$ is an integer counting the (non-negative) number of tails that have occurred.
Then we can begin with the states possible after the first three coin tosses, since these are uniformly distributed (due to "magic" initialization conditions). At time $t=3$:
$$ Pr(s(hhh,0)) = Pr(s(hht,1)) = Pr(s(hth,1)) = Pr(s(htt,2)) = \frac{1}{8} $$
$$ Pr(s(thh,1)) = Pr(s(tht,2)) = Pr(s(tth,2)) = Pr(s(ttt,3)) = \frac{1}{8} $$
Then each subsequent coin toss will again be a fifty-fifty affair except for the states $s(hhh,T)$ and $s(ttt,T)$, where "magic" causes it to be more likely ($75\%$) that the streak ends than that it continues.
All these probability transition rules could be expressed, without regard for the number of tosses possible or the number of tails allowed, by a semi-infinite Markov matrix:
$$ M = \begin{bmatrix}
A & B & 0 & 0 & \ldots \\
0 & A & B & 0 & \ldots \\
0 & 0 & A & B & \ldots \\
\vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} $$
where the states are "graded" as to the number $T$ of accumulated tails and:
$$ A = \begin{bmatrix}
0.25 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0.5 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0.5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0 \\
0.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0.5 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0.5 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0.75 & 0 \end{bmatrix} $$
accounts for coin tosses that produce "heads", and similarly matrix $B$ would account for coin tosses producing "tails":
$$ B = \begin{bmatrix}
0 & 0.75 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0.5 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0.5 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.5 \\
0 & 0.5 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0.5 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0.5 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.25 \end{bmatrix} $$
Let $v_3$ be the (row) vector with the probabilities of states distributed after the first three tosses (as above). Then the probability that there will be seven or more heads after ten tosses is:
$$ v_3 M^7 u^T $$
where $u$ is a (row) vector with ones in the first $32=8\cdot 4$ entries, and thus captures the total likelihood of no more than three tails at a given step.
Since we are only interested finally in the states where no more than three tails have appeared, we can truncate the vectors $v_3$ and $u$ and matrix $M$ accordingly. Thus let $v_*$ and $u_*$ be the truncations of $v_3$ and $u$ (resp.) to their first $32$ entries (corresponding to states with $T=0,1,2,3$) and similarly let $M_*$ be the leading principal $32\times 32$ submatrix of $M$:
$$ M_* = \begin{bmatrix}
A & B & 0 & 0 \\
0 & A & B & 0 \\
0 & 0 & A & B \\
0 & 0 & 0 & A \end{bmatrix} $$
Thus $v_3 M^7 u^T = v_* M_*^7 u_*^T$ also expresses the final probability (of seven or more heads in ten tosses).
We could make a further reduction from eight to six states in each graded block by adopting the idea expressed by the OP, of tracking only the current "run" of heads or tails, i.e. $h_k$ or $t_k$ in place of $L$, denoting $k=1,2,3$ immediately preceding heads or tails, respectively. This would bring down the dimension of the state vector from $32$ to $24$. Formally this would only affect the structure of our block submatrices $A,B$, which would become $6\times 6$ rather than size $8\times 8$.
Final Answer
I implemented the matrix multiplication in two spreadsheet environments separately, as a guard against misplaced cut-and-paste errors. The vector $v_*$ scales to integer entries when multiplied by $8$, and the matrix $M_*$ to integer entries when multiplied by $4$ (vector $u_*$ is already all ones). Thus the matrix product could be computed exactly with integer arithmetic.
Both the $8\times 8$ and $6\times 6$ block versions were implemented (on both platforms) as a guard against formula errors.
The result was probability of seven or more heads in ten tosses is $\frac{15808}{2^{17}} = \frac{13\cdot 19}{2^{11}} = 0.12060546875$.
