If
$x \in \ker BB^\ast \cap \ker A^\ast A, \tag 1$
then
$BB^\ast x = 0 = A^\ast A x; \tag 2$
then
$Cx = BB^\ast x + A^\ast A x = 0, \tag 3$
so
$x \in \ker C, \tag 4$
i.e.,
$\ker A^\ast A \cap \ker B B^\ast \subseteq \ker C, \tag 5$
as our OP yoyostein has affirmed. On the other hand, if (4) binds, then so does (3), and
$BB^\ast x + A^\ast Ax = 0, \tag 6$
from which we have
$x^\ast BB^\ast x + x^\ast A^\ast Ax = 0, \tag 7$
and since
$x^\ast B = (B^\ast x)^\ast, \; x^\ast A^\ast = (A x)^\ast, \tag 8$
(7) becomes
$(B^\ast x)^\ast B^\ast x + (A x)^\ast Ax = 0; \tag 9$
we now note that for any vector
$y = (y_1, y_2, \ldots, y_n)^T \tag{10}$
we have
$\Bbb R \ni y^\ast y = (y_1^\ast, y_2^\ast, \ldots, y_n^\ast)(y_1, y_2, \ldots, y_n)^T = \displaystyle \sum_1^n y_i^\ast y_i \ge 0; \tag{11}$
the fact that $y^\ast y \in \Bbb R$ may also be seen thusly:
$(y\ast y)^\ast = y^\ast (y^\ast)^\ast = y^\ast y; \tag{12}$
also, (11) shows that
$y^\ast y = 0 \Longleftrightarrow y = 0; \tag{13}$
now from (11)-(13) we find
$\Bbb R \ni (B^\ast x)^\ast B^\ast x, (A x)^\ast Ax \ge 0; \tag{14}$
and thus the only way (9) may bind is if
$(B^\ast x)^\ast B^\ast x, (A x)^\ast Ax = 0, \tag{15}$
which by (13) forces
$B^\ast x, Ax = 0; \tag{16}$
then
$BB^\ast x = A^\ast Ax = 0, \tag{17}$
so that
$x \in \ker BB^\ast \cap \ker A^\ast A, \tag{18}$
and finally we see that
$\ker C \subseteq \ker BB^\ast \cap A^\ast A; \tag{19}$
thus, combining (5) and (19),
$\ker C = \ker BB^\ast \cap A^\ast A. \tag{20}$