Great question, ultimately related to the definitions of derivatives and integrals using limits and the fundamental theorem of calculus. Not a mathematician, so things might be informal and a bit chaotic, but overall I hope my answer helps the intuition about these concepts.
Short answer:
Because the fundamental theorem of calculus makes this exact in the limit $dx\to0$.
Long but hopefully more intuitive answer:
Looking at the derivative
We start with your equation (1):
$$
\Phi_{x+dx}=\Phi_x+\left.\frac{\partial\Phi}{\partial x}\right|_xdx+o(dx^2).
$$
This relationship becomes exact in first order sense as we take the limit $dx\to0$:
$$
\lim_{dx\to0}\Phi_{x+dx}=\Phi_x+\lim_{dx\to0}\left.\frac{\partial\Phi}{\partial x}\right|_xdx.
$$
This might become clearer if we plug in the definition of the derivative:
$$
\lim_{dx\to0}\Phi_{x+dx}=\Phi_x+\lim_{dx\to0}\frac{\Phi_{x + dx}-\Phi_{x}}{dx}dx.
$$
In fact, we can lose the limit entirely here!
$$
\Phi_{x+dx}=\Phi_x+\frac{\Phi_{x + dx}-\Phi_{x}}{dx}dx=\Phi_x+\Phi_{x + dx}-\Phi_{x}=\Phi_{x + dx}.
$$
First- and second order expansions
For non-zero dx, a first order expansion approaches the exact function for smaller $dx$, and a higher-order expansion approaches that function faster (in a small region). In a sense, the "region" of confidence is smaller for a first order approximation. You already mention smoothness and texts on Taylor expansions elaborate on the region of convergence, requirements of the function, etc.
Exactness of first order expansion when taking the limit
Letting $dx$ approach zero, the higher order terms vanish quickly. In fact, $dx=10^{-3}$ gives $dx^2 = 10^{-6}$, $dx^3 = 10^{-9}$, etc., i.e. the order of magnitude of the difference between first and second (or third) order terms increases quadratically (or cubically, respectively). Therefore, we only need the first order term to determine $\Delta \Phi$ in a small region. If we weren't interested in changes, we could also use a constant value.
Letting taking the limit of $dx$ to zero could perhaps be more intuitively written as:
$$
\lim_{dx\to 0} dx = \lim_{n\to \infty} 10^{-n}
$$
and the difference in magnitude between first and second order terms scales with $dx$:
$$
\lim_{dx\to 0} \frac{dx}{dx^2} = \lim_{dx\to 0} \frac{1}{dx} \to \infty.
$$
(Technically, the limit doesn't exist I guess.)
Use in a derivative balance equation
The value of $\lim_{dx\to0}\Phi_{x+dx}$ is never used "raw". It is simply \Phi_{x} for a "smooth" function. Your resulting balance equation uses the gradients, or tangent line in the single variable case, which involves the derivative and thus dividing by the volume $V$. Somewhere along the way by doing that you implicitly divided by $dx$, resulting in the gradient/derivative of $\Phi$. Again, implicitly, you also used the limit $dx\to 0$ in your substitution, which allows you to strike out second order terms and remain exact.
Use in an integral equation
As an exercise, look at the integral form of the balance equation and the relation between the definition of the derivative and the integral as a Riemann sum. For simplicity, let's take a function $\Phi(x)$ on an interval $[a,b]$ and look at the arc length $L(\Phi)$ using the well-known arc-length formula:
The length of a small piece of the curve may be given as:
$$
L_{dx}(\Phi) = \sqrt{dx^2 + dy^2} = \sqrt{dx^2 + (\Phi(x+dx)-\Phi(x))^2} = dx\sqrt{1 + \left(\frac{\Phi(x+dx)-\Phi(x)}{dx}\right)^2}
$$
For a nonzero $dx$, this is a crude approximation, which could benefit from a second or higher order term. We can sum these segments up to get the length:
$$
L_{[a,b]}(\Phi) = dx\sum_{i=1}^n \sqrt{1 + \left(\frac{\Phi(x+dx)-\Phi(x)}{dx}\right)^2},
$$
for some appropriate $n$ and with $x=a+i*dx$. This sum approaches the true value as $dx$ approaches zero. Taking the limit transforms the finite difference into the derivative and the sum into the integral and results in the exact arc length:
$$
L_{[a,b]}(\Phi) = \int_a^b \sqrt{1 + \left(\frac{d\Phi(x)}{dx}\right)^2}dx,
$$
No need for a second order term!
