Notation: I use $X$ for an empty cell (no vampires or humans), $V$ for a vampire, $H$ for a human, and (in the process of solving the puzzle) $n$ for "either X or V" and $s$ for "either V or H".
A general note:
Any "line of sight" that goes directly from one number $p$ to another number $q$, bouncing off mirrors on the way, can be divided into (up to) three sections: (A) between $p$ and first mirror; (B) between mirrors (may be empty); (C) between last mirror and $q$. Anything in (B) counts equally for both $p$ and $q$, and so does any Human on the path. So the difference between $p$ and $q$ is precisely the difference between the number of Vampires in (A) and in (C).
Step by step
First the "obvious" deductions, from the 0s and a couple of easy 1s:
(Some other things are also easy to see right away: e.g. the first six cells of the bottom row must contain two $V$, one $H$, three $X$; and the first four cells of the third row must contain at least three $V$; but those aren't so easily shown with the notation.)
Now consider the $4,3$ at the top (fifth and sixth columns), the $5$ at the bottom, and the $5$ on the left.
All four cells that $4$ can see must be filled ($s$), and the two by the $3$ must be $H$.
By the "general note" at the beginning, all four cells above the bottom $5$ must be $V$, and the two cells past the mirror must be one $H$ and one $X$.
The left $5$ can only see six cells, past a mirror, and one of those is $n$, so all the others must be $H$. From the $6$ in the bottom left, the last one must be $V$.
Now the left $1$ in the fifth row is done, and we can fill in a bunch of $X$ and $n$ for that.
Now consider the $3,2$ path at the bottom right, and the paths around the top right triangle.
That $3$ (bottom row, third from the end) can only see three cells, so they must all be filled, and the two on the right must be $H$ and the bottom one $V$.
Now, comparing the $9,6$ path, and using the "general note" at the beginning, all four cells starting from the $9$ must be $V$, and the other two beside the $6$ are either $H$ or $X$. (The one next to the $6$ must be $X$ because of the $2$ below, and we can also fill in a lot of $n$ because of that $2$.)
Around the top right triangle, there is a path which starts from a $2$ (seventh on the right) and then splits to go to a $1$ (second on the right) and another $2$ on the top. On the lower part of the path (four cells along the right edge) there must be one $H$ and three $n$. The other part just has two cells; one is already $X$, so the other one must be $H$.
Now consider the $2,1$ path starting from the $2$ at the top of the right edge. There's already one $H$ on this path, so everything else must be $n$, which means the top right cell must be $V$, and the ones near the $1$ must be $X$.
Now consider the top row, and the paths around the bottom right triangle.
The $3$, seventh at the top, can only see Humans, so there must be three $H$ among the first six cells in the top row, and there's now only three which can be $H$. Now the other $3$, sixth at the top, can already see three $H$, so the $s$ below one of them must be $V$.
Also, consider again the $9,6$ path. The $9$ can see four $V$, so it must also see five $H$ beyond the mirrors, and there's now only five which can be $H$.
Now consider the $3$ in the bottom row second from the end. It can now see three $H$, so everything else on its path must be $n$. Then the $2$ on the right, fifth from the bottom, can see only $n$, one $H$, and the thing closest to it, which must therefore be $s$.
Let's also remember that the four central cells are empty, and consider again the $6$ on the right of the fifth row.
By the "general note" at the beginning, that $6$ must see at least two $V$ directly, since it's on a path with the $4$ above it. So we can place two $V$ in that row, and the cell next to the $4$ must be either $X$ or $H$.
Now I want to label some empty cells:
From the $6,4$ path, we know that exactly two of $a,b,d,e$ must be $H$. Assume it's not $e$; then it must be $b$ and exactly one of $a,d$, by the $3,3$ path around the lower left. Then the $3$ on the left of the seventh row can already see three things directly, so the $f$ cell is $X$ and the $c$ cell is $n$. But then the $7$ at the top can see at most six things, contradiction.
So the $e$ cell is $H$, which means the $s$ below it must be $V$.
In fact, the above contradiction can arise just from assuming that one of $a,d$ is $H$. So both of those must be $n$, and then the $6,4$ path tells us the $b$ cell is $H$.
Now the $7$ tells us that either $c$ is $H$ or $f$ is $s$; the $3,3$ path tells us it's not both, so that means both the $n$ on the $7$ path must be $V$.
Then the $4$ on the left of the second row is done, so we can fill in three $X$ there.
Now we're almost done, and deductions are falling like dominoes. I'm just making this pause for a breather between spoilertags.
Top of the first column: that $4$ can only be filled by a $V$ in the top left and an $H$ in the lower right.
Left of the first row: that $4$ is done, so we fill an $X$.
Top of the fourth column: that $3$ can only be filled by $V$ lower down.
Bottom of the second column: that $4$ can only be filled by $V$ just there and $H$ in the cell we called $c$ before.
Left of the bottom row: that $3$ is done, so we fill two $X$.
$3,3$ path in lower left: the left $3$ is done, so we fill two $X$, and then another $V$ for the bottom $3$.
Bottom of the sixth column: that $4$ can only be filled by $V$ above it.
Now we've filled all the $H$, and everything that can be deduced directly. I've emptied the $n$ squares so they stand out more clearly (there's nine of them):
Finally, since there's 22 Vampires in total, every blank square must be $X$ and we get the final solution.