This will be rather difficult.
Let's first of all suppose there are no problems, i.e. the functions you are interested in are infinitely differentiable and continuous in the range of values you are interested in.
Then, even with the simple example that you know the Maclaurin series of $f(x)$ and want the Taylor series of $f(x)$ about another point $a$, this is a cumbersome endeavour. Here is the procedure.
With the known Maclaurin series (you memorized the $m_k$)
$f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} x^k = \sum\limits_{k = 0}^{\infty} m_k x^k$
you require the Taylor series
$f_a(x) = \sum\limits_{k = 0}^{\infty} c_k (x-a)^k$
Now the coefficients are given by
$c_k = \frac{f^{(k)} (a)}{k!}$
which in turn you can get by differentiating your MacLaurin series and evaluate at $a$:
$c_k = \frac{f^{(k)} (a)}{k!} = \sum\limits_{n = k}^{\infty} \frac{n!}{k! (n-k)!} m_n a^{n-k}$
E.g.
$c_0 = \sum\limits_{n = 0}^{\infty} m_n a^{n}$
So you need to sum an infinite series to arrive at your new coefficients - big labour.
Another method is to identify the coefficients of all powers of $(x-a)$. To do so, write the MacLaurin series
$f(x) = \sum\limits_{k = 0}^{\infty} m_k ((x-a)+a)^k
= \sum\limits_{k = 0}^{\infty} m_k \sum_{j=0}^k {k \choose j} (x-a)^j (a)^{k-j}$
This leads to the same result.
Things get worse if you consider $g(f(x))$.
I guess this already answers your question: it's not a smooth technique to calculate the Taylor expansion from the known MacLaurin series.
A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$.
\begin{align*}
f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R
\end{align*}
Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence.
\begin{align*}
f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R
\end{align*}
We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$.
\begin{align*}
f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R}
\end{align*}
Substitution $u=2x$
We consider
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad 2x\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.
We obtain
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}
Substitution $u=x+1$
We consider
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x+1\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.
We obtain
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}
We also obtain
\begin{align*}
e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\
&=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\
&=e^{x+1}
\end{align*}
Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.
Best Answer
Def: A function $f$ is $o(x^n)$ for $x \to 0$ if $$\lim_{x \to 0} \frac{f(x)}{x^n} = 0$$
If $f$ is $o(x^n)$ for $x \to 0$, and if $m < n$, then $f$ is $o(x^m)$ too.
There is a theorem, due to Peano, that if $f(x) = P(x) + o(x^n)$, for $P(x)$ a polynomial of degree $n$, then $P(x)$ is the Taylor polynomial of order $n$ of $f$, around zero. Clearly the definition I gave right at the beginning can be generalized for another center, $x_0$.
I'll find the Taylor expansion around zero of $\ln (1 + x^2)$, of order $6$. It is known that: $$\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$$
Using $x^2$ instead of $x$, we get: $$\ln (1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} + o(x^6)$$ Since the remainder is $o(x^6)$, that is the polynomial desired, for Peano's theorem. Bigger examples might require that some terms are "swallowed" by the remainder $o(x^n)$, if the degree of said terms gets more than $n$. See that $o(x^n)$ is a notation, so things like $o(x^n) + o(x^n) = o(x^n)$ shouldn't bother you. At the end of the day, the $o(x^n)$ you have is not the same you began with. For example, $x^4 + x^5 + o(x^3) = o(x^3)$.
So, it is not simple as just making substitutions, you always have to pay attention to the remainder, or else you might get a polynomial, that is not the best approximation.