Remark: This will be one an attempt at a "fail fast" post and it is likely that a standard textbook contains answer to this question. The problem is that I am so new to the Laplace-Beltrami operator and metric tensors that I have not found a reference which I could apply in this situation.
Let $c > 0$ be a fixed and $\alpha,\beta$ be positive continuously differentiable functions on $x$ alone. Suppose that our ambient manifold is some compact subset $K\subset\mathbb{R}^2$ and our metric tensor is given by $g_{x,y}=(c + \alpha(x))dx^2 + \beta(x)dy^2$ at $(x,y)\in K$
I am looking to compute the image of a nice-enough function $f$ under the Laplace-Beltrami operator $\Delta_g=\frac{1}{\sqrt{|g|}}\partial_i\left(\sqrt{|g|}g^{ij}\partial_j f\right)$, where $|g|$ stands for the determinant of the matrix representation of $g$ and $g^{ij}$ the $ij$th entry of the inverse of the matrix representation of $g$.
I think that in our case the matrix representation of $M_{g_\cdot}$ is given by
$$M_{g_{x,y}}=\begin{bmatrix}(c + \alpha(x)) & 0\\\ 0 & \beta(x)\end{bmatrix}$$
with an inverse
$$M_{g_{x,y}}^{-1}=\begin{bmatrix}(c + \alpha(x))^{-1} & 0\\\ 0 & \beta(x)^{-1}\end{bmatrix}$$
therefore by denoting $\frac{\partial f}{\partial x}=f_x$, I believe that for nice enough $f$
\begin{align} \Delta_gf(x,y) &= \frac{1}{\sqrt{(c+\alpha(x))\beta(x)}}\bigg(\frac{\partial}{\partial x}\left(\sqrt{(c+\alpha(x))\beta(x)}(c + \alpha(x))^{-1}f_x(x,y)\right) \\ &+ \frac{\partial}{\partial y}\left(\sqrt{(c+\alpha(x))\beta(x)}\beta(x)^{-1}f_y(x,y)\right)\bigg) \\ &= \frac{f_{xx}(x,y)}{c + \alpha(x)} + f_x(x,y)\cdot \left(\frac{\partial}{\partial x}\frac{\sqrt{\beta(x)}}{\sqrt{c + \alpha(x)}}\right)(x,y) + \frac{f_{yy}(x,y)}{\beta(x)} \end{align}
where I have decided to omit the actual computation of $\frac{\partial}{\partial x}\frac{\sqrt{\beta(x)}}{\sqrt{c + \alpha(x)}}$. Besides this omission, do these computations seem correct?