I think that the missing link that connects the Fourier transform to the Taylor series expansion is Euler's formula, $e^{\jmath x}=\cos(x) +\jmath \sin(x)$. This celebrated formula establishes a relationship between trigonometric functions of real entities and exponential functions of complex (i.e. imaginary) entities. In doing so, it establishes a fundamental connection with the Fourier transform, which is essentially trigonometric in nature. In fact, the Fourier transform for a function $f(t)$ is, by definition, a Fourier series of $f(t)$ interpreted as $t\rightarrow \infty$, and a Fourier series is, by definition, a linear summation of sin and cos functions. The Taylor Series comes into play in the derivation of Euler's formula.

As previously mentioned, Euler's formula states $e^{\jmath x}=\cos(x) +\jmath \sin(x)$. Hence, it's acceptable to conceptually superimpose the conventional (x, y) unit circle and the real-complex plane, as they both portray the polar Eulerian expression on the continuous interval from $0$ to $2\pi$. We therefore, evaluating Euler's formula at $x=2\pi$, arrive at
$$e^{2\pi*\jmath \theta } = {\cos(}2\pi\theta)+j*{\sin}(2\pi\theta)$$
This expression is intrinsically intertwined with the nature of a Fourier transform because Fourier transforms aim to convert a function from the time domain to the frequency domain. To do so, we decompose periodic functions into simple, linear sums of sin and cos, and allow this new function to approach infinity.

So, where does the Taylor series fit into all this? We use the Taylor series expansion (or more precisely, the McLauren series expansion) to obtain Euler's formula. Here's the proof (or derivative, depending on your perspective):

*Proposition*: For any complex number $z$, where $z=x+\jmath y=x+\sqrt{-1}*y$, i.e. $x=Re \{z \}\ $ and $y=Im \{z \}\ $, it can be said that, for real values $x$,

**$$ e^z=e^{j\pi} = \cos(x)+\jmath \sin(x)$$**

*Lemma 1*: The Taylor series expansion of an infinitely differentiable function $f(x)$ in the neighborhood $a$ is defined
$$f(x)=\sum_{k=0}^{\infty} \frac{f^k(a)}{k!}(x-a)^k$$
We'll say that we're working in the neighborhood $a$ surrounding $\theta$. So, when $\theta = 0$, we get a special case of the Taylor series, called the McLauren series. The expansion of the Taylor (or McLauren) series for $\sin(x)$ and $\cos(x)$ are
$$ \sin (\theta)=\sum_{k=0}^{\infty} \frac{d^{k} \sin(\theta)}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} ...$$
$$ \cos (\theta)=\sum_{k=0}^{\infty} \frac{d^{k} \cos(\theta)}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} ...$$
This is not a surprise, as we know that sine and cosine are odd and even functions, respectively. We've now accounted for the trigonometric portions of $e^z$, but have not yet addressed the complex component.

*Lemma 2*: We apply the previously defined Taylor (more specifically McLauren) series expansion to the function $e^z$ when $z=\jmath x$, and we get
$$e^{\jmath x} =\sum_{k=0}^{\infty} \frac{d^{k} e^{\jmath x}}{d\theta^k} \mid _{\theta=0} = \sum_{k=0}^{\infty}\frac{\theta^k}{k!} = 1 + \jmath x -
$$

Comparing the right-hand terms from the Taylor series expansions of $ \cos (\theta)$, $\sin (\theta)$, and $e^{\jmath \theta}$ performed in Lemmas 1 and 2, we see that summing the expansions from Lemma 1, we arrive, by the procedure I'm too lazy to type out in LaTeX but not too lazy to explain via my good friends Google and krotz, at
$e^{\jmath x}=\cos(x) +\jmath \sin(x)$. [Q.E.D.]

Now that we've proven Euler's formula, we can evaluate it at $x=2\pi$ to acquire

$$e^{2\pi*\jmath \theta } = {\cos(}2\pi\theta)+j*{\sin}(2\pi\theta)$$

Due to the characteristics of the sin and cos functions, it is possible to use simple integration to recover the amplitude of each sin and cos wave represented in a Fourier transform (similar to the reverse of the above proof). In an overwhelming majority of cases, it's highly useful to select Euler's formula as the function to integrate over. Because Euler's formula and the Fourier transform are both (at least, in part) fundamentally trigonometric in nature, the use of Euler's formula greatly simplifies most of the real portions of the Fourier analyses. Additionally, for the complex case, frequencies can be represented inverse-temporally using a combination of Euler's formula and the Fourier series expansion.

So, it's a bit messy and convoluted (etymologically, not integrally), but it really boils down to the fact that the Taylor (or McLauren) series, the Fourier series and transform, and Euler's formula all relate a trigonometrically
The differences between the three arise by nature of application. Taylor series are used to represent functions as infinite sums of their derivatives. Fourier series and transforms are used in linear systems &/or differential equations to convert signals or DEs from the time to frequency domain. Euler's formula is used to relate trigonometric and complex exponential (complexponential?!) funcions, and is also a formula that, when evaluated at $x=\pi$, yields Euler's identity, $e^{\jmath \pi}+1=0$, an equation so austerely eloquent and aesthetically arousing that I'd be down to stare at all day.