Orthogonal decomposition in indefinite inner products Consider a finite dimensional vector space V which equipped with a bilinear form B (inner product) not necessarily positive definite. Orthogonality and orthogonal complements with respect to B
are defined as usual. A subset $E\subseteq V$ is called $\textit{orthonormal}$ if
$B(e,f)=\pm\delta_{ef}$, for all $e,f\in E$.
The bilinear form $B$ is called nondegenerate if there is no nonzero $v\in V$ which is orthogonal to each vector $w\in V$ (including itself), i.e. if $V^\perp=\{0\}$. Assume that this is the case. Then it can be shown that $V$ has an orthonormal basis. Conversely, if $V$ has an orthonormal basis, then it is easy to see that $B$ must be nondegenerate.
Now consider a subspace $H\subseteq V$. It is not true in general that $H\cap H^\perp=\{0\}$
(e.g. $H=span(v)$ where $v\in V$ with $B(v,v)=0$). In this case we do not have the direct sum decomposition $V=H\oplus H^\perp$ and the orthogonal projection $P_H:V\to H$ is not defined.
This is pathological and needs to be ruled out. It is easy to see that the following are equivalent:
(i) $H\cap H^\perp=\{0\}$, (ii) the restriction of $B$ to $H$ is nondegenerate, and
(iii) $H$ has an orthonormal basis.
(I) But is it always true that $V=H+H^\perp$?
Now consider a smooth connected manifold $M$ embedded in $V$. Then the tangent spaces $T_x(M)$ are all subspaces of $V$.
(II) Are there any natural conditions that ensure that the restriction of $B$ to $T_x(M)$ is nondegenerate, for all points $x\in M$ (other than $B$ is positive definite).
(III) Assume that the tangent spaces $T_x(M)$ all have orthonormal bases. Must the signature of the quadratic form $B$ be the same on all tangent spaces?
(IV) Are there any natural conditions that ensure that $V=T_x(M)\oplus T_x(M)^\perp$, for all points $x\in M$.
(V) If this is true for one point $x\in M$, does it follow that it is true for all others too?
I am writing myself some notes on manifold theory assuming that $M$ is embedded in such $V$ to simplify the exposition and with the intention of using the Clifford algebra $G(V)$. The above points come up all the time.
 A: I won't give all the details since I believe it's better for the inquirer to try and figure them out, but if any doubt persists, I'll be glad to help some more in the comments.
I'll assume that $V$ is a finite-dimensional $\Bbb R$-vector space.
(I) Just recall that $\dim V=\dim H+\dim H^{\perp}$ if the form is non-degenerate.
Thus, if $H\cap H^{\perp}=0$, then yes $H+H^{\perp}=V$. Actually, as you noticed, $V=H\oplus H^{\perp}$ in that case.
Otherwise, if $\dim(H\cap H^{\perp})=d\ne0$, then $\dim(H+H^{\perp})=\dim V-d\ne\dim V$, so $H+H^{\perp}\ne V$.
Moreover, if $H_1,H_2$ are subspaces of $V$ such that $H_1\cap H_1^{\perp}=H_2\cap H_2^{\perp}$, then it turns out that $H_1+H_1^{\perp}=H_2+H_2^{\perp}$. Indeed, this follows from $(H\cap H^{\perp})^{\perp}=H+H^{\perp}$.
(II) Note that the tangent spaces to $M$ are not necessarily subspaces of $V$, since they not necessarily pass through the origin of $V$. They are $k$-planes in the affine space $\Bbb A$ associated to $V$, where $k=\dim M$.
However, it's true that the embedding $M\rightarrow V$ induces a smooth map $\tau:M\rightarrow{\rm Gr}(k,V)$, where ${\rm Gr}(k,V)$ is the Grassmannian of $k$-subspaces of $V$. Indeed, for each $x\in M$, there's a unique affine translation $t_x:\Bbb A\rightarrow\Bbb A$ such that $t_x(x)=0$, and $t_x$ depends smoothly on $x$. Thus, we can define $\tau(x):=t_x(T_xM)$.
(Isn't it interesting that this construction relies on the fact that the affine translations form a distinguished subgroup of the affine group? Actually, I believe one might say that all of linear algebra ultimately amounts to that fact!)
Now, the desired condition can be spelled out as

$\tau(x)\cap\tau(x)^{\perp}=0$ for all $x\in M$.

Note that the 'restriction of $B$ to $T_xM$' only makes sense via some correspondence as described above (are there other choices?): one must pull the restriction of $B$ to $t_x(T_xM)$ back to $T_xM$ via $t_x$.
(III) Since $M$ is connected, and assuming that the condition above holds, then yes, all of them must have the same signature. In order to understand that, it's useful to define ${\rm Gr}^0(k,V):=\{H\in{\rm Gr}(k,V)\mid H\cap H^{\perp}=0\}$ and note that ${\rm Gr}^0(k,V)$ is a disjoint union of connected components indexed by signature. Now, it suffices to note that the signature of $\tau(x)\in{\rm Gr}(k,V)$ depends continuously on $x\in M$.
(IV) Same condition as in (II), just recalling that the summand shouldn't be considered $T_xM$ itself, but $t_x(T_xM)$, otherwise the direct sum doesn't make sense.
(V) No: if $\tau(x)\cap\tau(x)^{\perp}=0$ for some point $x\in M$, it doesn't follow that the same holds for all $x\in M$.
Just an additional remark: I disagree that the case $H\cap H^{\perp}\ne0$ is 'pathological'. It depends on what you're interested in! For example, the existence of a conformal structure on the absolute of the hyperbolic space turns out to be related to this case, and it's a very nice feature of these spaces. But if you said that this case demands some special care to be dealt with, then I would agree, that's for sure. =]
