Let $(M, g) = (\mathbb{R}^n , ds^2 = dr^2 +f^2(r)d\theta^2)$ where $d\theta^2$ is the induced Riemannian metric from $\mathbb{S}^{n-1}$. Determine the Riemannian volume form of $M$.
The volume form is in local coordinates given by $$\omega = \sqrt |g| dx^1 \wedge \dots \wedge dx^n$$ so the whole problem reduces to finding $\sqrt{|g|}$ where $|g|$ is the absolute value of the determinant of the metric tensor.
Now in this case $g = ds^2$ so we are trying to find the matrix representation for $ds^2$. This is where I'm stuck I think we should end up with a matrix $g_{ij}$, but since this depends on $d\theta^2$ also I don't know how to find this matrix. Any help would be appreciated.
Here is a coordinate-free proof. Note that usual Euclidean coordinates are not suitable for this problem due to the spherical symmetry, and computing the determinant of the metric in these coordinates will be painful. $\DeclareMathOperator{\d}{d} \newcommand{\dx}{\d\!} \DeclareMathOperator{\vol}{vol}$ The proof I give relies on the fact that if $\{\alpha^1,\ldots,\alpha^n\}$ is an oriented orthonormal coframe, then the Riemannian volume form is $\alpha^1\wedge\cdots\wedge\alpha^n$ (this follows straightforwardly from the fact that the Riemannian volume form is the unique volume form that is $1$ on any oriented orthonormal frame).
Let $\{e_1,\ldots,e_{n-1}\}$ be an oriented orthonormal frame on $S^{n-1}$, and $\alpha^i = g_{S^{n-1}}(\cdot,e_i)$ be the dual $1$-forms with respect to the round metric (that you call $\dx\theta^2)$. The volume form of $S^{n-1}$ then reads $$ \dx\vol_{S^{n-1}} = \alpha^1\wedge \cdots \wedge \alpha^{n-1}. $$ From the form of the metric you give, it is clear that $\{\partial_r,\frac{1}{f}e_1,\ldots,\frac{1}{f}e_{n-1}\}$ is an orthonormal frame for $(\Bbb R^n,\dx s^2)$. Up to a changing the orientation, let's assume that it is also oriented. The dual coframe is $\{\dx r,f\alpha^1,\ldots,f\alpha^{n-1}\}$, from which is derived that the volume form is \begin{align} \dx\vol_{\dx s^2} &= \dx r\wedge(f\alpha^1)\wedge\cdots\wedge(f\alpha^{n-1}) \\ &= f^{n-1}\dx r\wedge\alpha^1\wedge\cdots\wedge\alpha^{n-1} \\ &= f^{n-1} \dx r\wedge \dx\vol_{S^{n-1}} \\ &= f^{n-1} \dx\vol_{eucl}, \end{align} where $\dx\vol_{eucl} = \dx r\wedge \dx\vol_{S^{n-1}}= \dx x^1\wedge\cdots\wedge \dx x^n$ is the usual Euclidean volume form (indeed, it is $1$ on the oriented Euclidean frame $\{\partial_r,e_1,\ldots,e_{n-1}\}$).
A proof in spherical coordinates would basically lead to the same computations, considering $\{y^1,\ldots,y^{n-1}\}$ coordinates on the sphere, so that $\{r,y^1,\ldots,y^{n-1}\}$ are coordinates on $\Bbb R^n$.
Here is a shorter way of basically doing the same thing. It relies on the fact that if one can write $g = \sum_{j=1}^n (\alpha^j)^2$ where $\alpha^j$ are $1$-forms, then $\{\alpha^1,\ldots,\alpha^n\}$ is an orthonormal coframe (and conversely), and the Riemannian volume form is then $\pm\alpha^1\wedge\cdots\wedge\alpha^n$ (depending on the orientation). Let $\{\alpha^1,\ldots,\alpha^{n-1}\}$ be an orthonormal frame of the round metric $g_{S^{n-1}}$. Then $g_{S^{n-1}} = \sum_{j=1}^{n-1} (\alpha^j)^2$. It follows that your metric reads $$ \dx s^2 = \dx r^2 + f^2\sum_{j=1}^{n-1} (\alpha^j)^2 = \dx r^2 + \sum_{j=1}^{n-1} (f\alpha^j)^2. $$ Hence, $\{\dx r,f\alpha^1,\ldots,f\alpha^{n-1}\}$ is an orthonormal coframe for $\dx s^2$. The rest of the proof is similar to the one above.
If you really want a proof using coordinates, here is one. Let $\{y^1,\cdots,y^{n-1}\}$ be coordinates on $S^{n-1}$, so that $\{r,y^1,\ldots,y^{n-1}\}$ are coordinates on $\Bbb R^n$. The associated frame is $\{\partial_r,\partial_1,\ldots,\partial_{n-1}\}$. Let's consider the Euclidean metric first. This frame has the property that $\partial_r \perp \partial_j$ for all $j\in \{1,\ldots,n-1\}$, and that $\|\partial_r\|=1$, and the matrix of the Euclidean metric has this the form $$ \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & & & \\ \vdots & & M & \\ 0 & &&& \end{pmatrix} $$ where $M$ is the matrix $M_{ij} = \langle \partial_i,\partial_j\rangle$. Hence, the Euclidean volume form is $$ \dx x^1\wedge\cdots \wedge \dx x^n = \sqrt{|\det M|} \dx r \wedge \dx y^1\wedge \cdots \wedge \dx y^{n-1}. $$ Now, consider the new metric $ds^2$. The matrix of this latter metric in the coordinates $\{r,y^1,\ldots,y^{n-1}\}$ is now $$ \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & & & \\ \vdots & & M' & \\ 0 & &&& \end{pmatrix} $$ with $M'$ the matrix $M'_{ij} = f^2 \langle \partial_i,\partial_j\rangle = f^2M_{ij}$. Hence, $M' = f^2M$. It follows that \begin{align} \dx \vol_{\dx s^2} &= \sqrt{|\det M'|} \dx r \wedge \dx y^1\wedge \cdots \dx y^{n-1} \\ &= \sqrt{|\det f^2M|} \dx r \wedge \dx y^1\wedge \cdots \dx y^{n-1} \\ &= \sqrt{f^{2(n-1)}} \sqrt{|\det M|}\dx r \wedge \dx y^1\wedge \cdots \dx y^{n-1} \\ &= f^{n-1} \sqrt{|\det M|}\dx r \wedge \dx y^1\wedge \cdots \dx y^{n-1} \\ &= f^{n-1} \dx x^1\wedge \cdots \dx x^n. \end{align}