$
\def\h{\odot}
\def\o{{\tt1}}
\def\bR#1{\big(#1\big)}
\def\BR#1{\Big[#1\Big]}
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\Diag#1{\op{Diag}\LR{#1}}
\def\tr#1{\op{tr}\LR{#1}}
\def\frob#1{\left\| #1 \right\|_F}
\def\qiq{\quad\implies\quad}
\def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}}
\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
\def\gradLR#1#2{\LR{\grad{#1}{#2}}}
$
Here is a numerical counter-example to your first formula using random, non-commuting, $2\times 2$, SPD matrices.
$$
A = \m{
29 & 57 \\
57 & 117 \\
},\,\,\,\,
X = \m{
20 & 44 \\
44 & 100 \\
},\,\,\,\,
dX = \m{
4 & 5 \\
5 & 18 \\
}\times 10^{-4}
$$
Let's estimate $df$ using your formula versus a direct calculation.
$$\eqalign{
f(X) &= \tr{\log(X)\cdot A} \\
f(X+dX)-f(X) &= 0.002849328 \\
\tr{AX^{-1}dX} &= 0.000975000 \\
\Delta &= 65.8\% \cr
}$$
To follow up on @lynn's comment, let's see what happens if the matrices commute. The simplest way to ensure that is to set $A=X$ and repeat the calculation.
$$\eqalign{
f(X+dX)-f(X) &= 0.00219992 \\
\tr{AX^{-1}dX}= \tr{dX} &= 0.00220000 \
\Delta &= 0.004\% \\
\\
}$$
Update
Here is a simple non-numerical example of what goes wrong when the matrices don't commute.
$$\eqalign{
f &= \tr{X^3A} \\
df &= \tr{X^2\,dX\,A + X\,dX\,XA + dX\,X^2A} \\
\grad fX &= X^2A + XAX + AX^2 \\
}$$
A rather ugly and complicated result for such a simple function. However if $(A,X)$ commute, then you can combine terms to obtain
$$\grad fX = 3X^2A$$
Now imagine expanding a matrix function as a Taylor series, and then taking its derivative term-by-term. Each term $X^k$ will explode into $k$ distinct terms and you'll end up with a horrible mess. But you could do it.
However, for the $\log$ function, you can't even write down a Taylor series because it's singular at zero.
Update 2
For SPD matrices, the $\sf Daleckii$-$\sf Krein\;Theorem$ yields a closed-form solution.
First, calculate the Eigenvalue Decomposition
$$\eqalign{
\def\b{\beta}
X &= QBQ^T,\qquad I=Q^TQ,\;B=\Diag{\b_k} \\
}$$
Applying a generic function $h(z)$ and its derivative $h'(x)$ to $X$ yields
$$\eqalign{
H_x &= h(X),\quad &H_b = h(B) \qiq &H_x = Q\,H_b\,Q^T \\
H_x' &= h'(X),\quad &H_b' = h'(B) \qiq &H_x' = Q\,H_b'\,Q^T \\
}$$
Since $B$ is diagonal, its functions are very easy to evaluate.
According to the DK Theorem the differential of this function is
$$\eqalign{
Z &= \zeta(BJ-JB), \quad R = \fracLR{H_bJ-JH_b+ZH_b'}{BJ-JB+Z} \\
dH_x &= Q\BR{R\h\LR{Q^TdX\,Q}}Q^T \\
}$$
where $J$ is the all-ones matrix,
$\LR{F\h G}$ denotes elementwise multiplication,
$\fracLR FG$ denotes elementwise division,
and $\zeta$ is an elementwise zero-indicator function, i.e.
$$\eqalign{
\zeta\!\LR{\m{-2 & \c0 & 3\\9 & -5 & \c0}}
= \m{0 & \c\o & 0\\0 & 0 & \c\o}
}$$
In the current problem
$$\eqalign{
h(x) &= \log(x), \qquad h'(x) = x^{-1} \\
}$$
The final piece of notation that we need is the Frobenius $(:)$ product
$$\eqalign{
F:G &= \sum_{i=1}^m\sum_{j=1}^n F_{ij}G_{ij} \;=\; \tr{F^TG} \\
G:G &= \frob{G}^2 \\
F:G &= G:F \;=\; G^T:F^T \\
\LR{PQ}:G &= P:\LR{GQ^T} \;=\; Q:\LR{P^TG} \\
\LR{E\h F}:G &= E:\LR{F\h G} \\
}$$
Putting this all together
$$\eqalign{
f &= A:H_x \\
df
&= A:dH_x \\
&= A:\LR{Q\BR{R\h\LR{Q^TdX\,Q}}Q^T} \\
&= \LR{Q\BR{R\h\LR{Q^TAQ}}Q^T}:dX \\
\grad fX &= Q\BR{R\h\LR{Q^TAQ}}Q^T \\
}$$
