Let $f : M\to \mathbb R$ be a smooth function, where $M$ is a smooth (infinite-dimensional, if you like) manifold. Pick a point $m\in M$. Then the value $f(m) = f^{(0)} \in \mathbb R$ is well-defined, of course, and also $f^{(1)}(m) = {\rm d}f_m$ is well-defined as a linear map ${\rm T}_mM \to \mathbb R$, i.e. as an element in ${\rm T}^*_m M$. In general, the higher derivatives are not well-defined, even if you choose a metric, although they are if you choose coordinates. What is well-defined is the "$k$-jet" of the function for each $k$; the $k$-jet of $f$ takes values in a vector bundle over $M$ that's locally isomorphic to $\bigoplus_{j=0}^k (\operatorname{Sym}^j( {\rm T}_m M))^*$, but choosing such an isomorphism essentially requires choosing local coordinates.

(Actually, more is true: the $k$-jet bundle surjects onto the $(k-1)$-jet bundle, so that there is a well-defined $\infty$-jet bundle; these surjections are compatible with the local trivializations, so that a trivialization identifies the $\infty$-jet bundle with $(\operatorname{Sym}^\bullet({\rm T}_mM))^*$. Naively you would define the Taylor series as living here. The system of surjections defines a filtration on the $\infty$-jet bundle, and $(\operatorname{Sym}^\bullet({\rm T}_mM))^*$ is the corresponding associated graded bundle.)

But there is one situation in which you can define a symmetric tensor $f^{(k)}(m) : ({\rm T}_mM)^{\otimes k} \to \mathbb R$. Namely, if all $f^{(1)}(m), \dots, f^{(k-1)}(m)$ vanish, then $f^{(k)}(m)$ is well-defined. This is because under changes of coordinates, the coordinate-dependent term $f^{(k)}(m) \in (\operatorname{Sym}^k({\rm T}_m M))^*$ changes by some term depending linearly on $\{f^{(1)}(m), \dots, f^{(k-1)}(m)\}$ (and polynomially on the derivatives of the change-of-coordinates function).

This is the situation for the Hessian. If $m$ is a critical point for $f$, then $f^{(1)}(m) = {\rm d}f_m = 0$, and so the second derivative is well-defined as a symmetric map $H = f^{(2)}(m) : ({\rm T}_mM)^{\otimes 2} \to \mathbb R$.

In any case, if $V$ is any vector space, and $\phi: V^{\otimes 2} \to \mathbb R$ any symmetric linear map, then the *Morse index* of $\phi$ is defined as the dimension (if it is finite, or $\infty$ otherwise) of any maximal subspace $W\subseteq V$ so that $\phi\rvert_{W\otimes W}$ is negative-definite. It is a straightforward lemma that this number (or $\infty$) does not depend on which maximal subspace you use. Of course, you can also define a "Morse coindex" as the Morse index of $-\phi$, and finally one can write down the degeneracy of $f$ at $m$ as the dimensional of the kernel of $\phi$ thought of as a map $\phi : V \to V^*$.

Point being, there are no other choices necessary to go: smooth function $f: M\to \mathbb R$ and choice $m\in M$ of critical point $\Rightarrow$ Hessian of $f^{(2)}(m) : ({\rm T}_mM)^{\otimes 2} \to \mathbb R$ $\Rightarrow$ Morse index of $f^{(2)}(m)$. At non-critical points, you cannot do the first step.

Finally, let me comment on the use of a metric. For applications of Morse theory to the topology of manifolds, picking a metric is important, as it lets you convert the differential one-form ${\rm d}f$ into a vector field and thus consider the corresponding flow lines, etc. From a topological point of view, what particular metric you use doesn't matter much: the space of metrics is contractible. But it is important; just not for the definition of Morse index.

Recall that a metric $g$ determines for each $m\in M$ a linear map $\operatorname{Sym}^2 {\rm T}_m M \to \mathbb R$, and hence a linear map $g_m: {\rm T}_m M \to {\rm T}^*_m M$. Being a metric means that this linear map is an injection (and hence an iso in finite-dimensions). So you should try to write down an inverse $g^{-1}_m: {\rm T}_m^* M \to {\rm T}_m M$. You can do this in finite-dimensions; in infinite dimensions the inverse is not everywhere-defined. Then you can define $\nabla f_m = {\rm d}f_m \, g^{-1}_m$ and, if you want, the Hessian to be the matrix $f^{(2)}_m g^{-1}_m$. Since $g$ is positive-definite, so is $g^{-1}$, and so the dimensions of the kernel and positive and negative eigenspaces of $f^{(2)}_m g^{-1}_m$ are the same as for $f^{(2)}$.