Yes, a neural network plus loss function can be viewed as a function composition as you have written, and back propagation is just the chain function repeated. Your equations $L = f(g(h(\dots u(v(\dots))))$ and $\frac{\partial L}{\partial v} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}\dots\frac{\partial u}{\partial v}$ are useful for high level intuition.
At some point you need to look at the actual forms of the functions, and that introduces a bit more complexity. For instance, the loss function for a mini-batch can be expressed in terms of the output. This is the equivalent of $L = f(g)$ for a batch or mini-batch with mean squared error loss, where I am using $J$ (for cost) in place of $L$ and the output of the neural network $\hat{y}$ in place of $g$:
$$J = \frac{1}{2N}\sum_{i=1}^{N}(\hat{y}_i - y_i)^2$$
The gradient of $J$ with respect to $\hat{y}$ is equivalent to your first part $\frac{\partial f}{\partial g}$:
$$\nabla_{\hat{y}} J = \frac{1}{N}\sum_{i=1}^{N}(\hat{y}_i - y_i)$$
Many of the functions in a neural network involve sums over terms. They can be expressed as vector and matrix operations, which can make them look simpler, but you still need to have code somewhere that works through all the elements.
There is one thing that the function composition view does not show well. The gradient you want to calculate is $\nabla_{\theta} J$, where $\theta$ represents all the parameters of the neural network (weights and biases). The parameters in each layer are end points of back propagation - they are not functions of anything else, and the chain rule has to stop with them. That means you have a series of "dead ends" - or possibly another way of thinking about it would be that your $\frac{\partial L}{\partial v} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}\dots\frac{\partial u}{\partial v}$ is the "trunk" of the algorithm that links layers together, and every couple of steps there is a "branch" to calculate the gradient of the parameters that you want to change.
More concretely, if you have weight parameters in each layer noted as $W^{(n)}$, and two functions for each layer (the sum over weights times inputs, and an activation function) then your example ends up looking like this progression:
$$\frac{\partial L}{\partial W^{(n)}} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}\frac{\partial h}{\partial W^{(n)}}$$
$$\frac{\partial L}{\partial W^{(n-1)}} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}\frac{\partial h}{\partial i}\frac{\partial i}{\partial j} \frac{\partial j}{\partial W^{(n-1)}}$$
$$\frac{\partial L}{\partial W^{(n-2)}} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}\frac{\partial h}{\partial i}\frac{\partial i}{\partial j}\frac{\partial j}{\partial k}\frac{\partial k}{\partial l} \frac{\partial l}{\partial W^{(n-2)}}$$
. . . this is the same idea but showing the goal of calculating $\nabla_{W} L$ which doesn't fit into a single chain of gradients. Notice the last term on each line does not appear in the next line. However, you can keep all the terms prior to that and re-use them in the next line - this matches the layer-by-layer calculations in many implementations of back propagation.
