Let $A_k$ be the leading principal $k\times k$ submatrix of $A$. Clearly, $A_1=S_1S_1^T$ where $S_1=\sqrt{A_1}$ is a real $1\times1$ lower triangular matrix. Now suppose that for some $k$, $A_{k-1}=S_{k-1}S_{k-1}^T$ for some real lower triangular matrix $S_{k-1}$. Since $A_{k-1}$ is positive definite, $S_{k-1}$ is nonsingular. Also, as $u^TA_ku>0$ for every nonzero vector $u$, if we write $A_k=\pmatrix{A_{k-1}&v_k\\ v_k^T&a_k}$ and put $u=(-v_k^TA_{k-1}^{-1},\,1)^T$, we obtain $a_k-v_k^TA_{k-1}^{-1}v_k>0$. Therefore the equation
$$
\pmatrix{A_{k-1}&v_k\\ v_k^T&a_k}
=\pmatrix{S_{k-1}&0\\ x^T&s}\pmatrix{S_{k-1}^T&x\\ 0&s}\tag{1}
$$
has a unique solution $x=S_{k-1}^{-1}v_k,\,s=\sqrt{a_k-x^Tx}=\sqrt{a_k-v_k^TA_{k-1}^{-1}v_k}$. This means $A_k=S_kS_k^T$ for some real lower triangular matrix $S_k$. So, if we start from $A_1$ and keep solving $(1)$ for $k=2,3,\ldots$, we see that $A=SS^T$ for some real lower triangular matrix $S$.
Let's use a naming convention where matrices and vectors are denoted
by upper and lower case Latin letters, respectively. Further, the symbol $\odot$ will denote the Hadamard product and $\otimes$ the Kronecker product.
For ease of typing, use $\{S,A,P\}$ instead of
$\,\{\Sigma,{\large\Lambda},{\cal P}\}\,$
and $\,X=f(\Sigma)\,$.
Then rewrite the problem using these conventions.
$$\eqalign{
S &= XX^T,\quad
A = I\odot X,\quad
V=A^{-1} \\
P &= VSV + A - I \\
}$$
Each of these matrices (except for $X$) is symmetric, and $(A,V,I)$ are diagonal.
Apply the vec operation ($K$ denotes the Commutation matrix)
$$\eqalign{
y &= {\rm vec}(I) \\
x &= {\rm vec}(X) \quad\implies\quad{\rm vec}(X^T) &\doteq Kx \\
a &= y \odot x
\;=\; {\rm Diag}(y)\,x
&\doteq Yx \\
da &= Y\,dx \\
\\
s &= (I\otimes X)Kx \;=\; (X\otimes I)\,x \\
ds &= \Big((I\otimes X)K+(X\otimes I)\Big)\,dx &\doteq N\,dx\\
\\
p &= (V\otimes V)s + a-y &\doteq Bs + a-y \\
&= (VS\otimes I)v + a-y &\doteq Hv + a-y \\
&= (I\otimes VS)v + a-y &\doteq Jv + a-y \\
}$$
Finally, calculate the differentials of $v$ and $p$
$$\eqalign{
dv &= {\rm vec}(-V\,dA\,V) = -(V\otimes V)\,da \\
&= -B\,da \;=\; -BY\,dx \\
\\
dp &= da + B\,ds + H\,dv + J\,dv \\
&= Y\,dx + BN\,dx - (H+J)BY\,dx \\
&= \Big(Y + BN - (H+J)BY\Big)\,dx \\
}$$
and the gradient with respect to $x$
$$\eqalign{
\frac{\partial p}{\partial x}
&= Y + BN - (H+J)BY \\
&= Y
+ (V\otimes V)\Big((I\otimes X)K+(X\otimes I)\Big)
- (VS\otimes I + I\otimes VS)(V\otimes V)Y \\
&= Y + (V\otimes VX)K+(VX\otimes V)
- \big(VSV\otimes V + V\otimes VSV\big)Y \\
}$$
Best Answer
We can use the chain rule,
$$ \begin{align} \frac{\partial \text{vech} (\mathbf{L} \mathbf{D} \mathbf{L}^\top)}{\partial \text{vech}(\mathbf{\Sigma})^\top} &= \frac{\partial \text{vech} (\mathbf{L} \mathbf{D} \mathbf{L}^\top)}{\partial \text{vech}(\mathbf{L})^\top} \frac{\partial \text{vech} (\mathbf{L})}{\partial \text{vech}(\mathbf{\Sigma})^\top} \\ &= \mathbf{G}^{+} \left( \mathbf{L} \mathbf{D} \otimes \mathbf{I} \right) \mathbf{F}^{\top} \left( \mathbf{G}^{+} \left(\mathbf{L} \otimes \mathbf{I}\right) \mathbf{F}^{\top} \right)^{-1}. \end{align} $$
To see this, let $\mathbf{G}$ be the duplication matrix, $\mathbf{G}^{+}$ its Moore-Penrose inverse, $\mathbf{F}$ the canonical elimination matrix (which consists of 0's and 1's only) and $\mathbf{K}$ the commutation matrix. Then $$ \begin{align} \mathrm{d} \mathrm{vech}\left( \mathbf{L} \mathbf{D} \mathbf{L}^{\top} \right) &= \mathbf{G}^{+} \, \, \mathrm{d} \mathrm{vec} \left( \mathbf{L} \mathbf{D} \mathbf{L}^{\top} \right) \\ &= \mathbf{G}^{+} \left( \mathrm{vec}\left( \mathrm{d} \mathbf{L} \mathbf{D} \mathbf{L}^{\top} \right) + \mathrm{vec}\left( \mathbf{L} \mathbf{D} \mathrm{d} \mathbf{L}^{\top} \right) \right) \\ &= \mathbf{G}^{+}\left(\mathbf{I} + \mathbf{K}_{pp}\right) \mathrm{vec}\left( \mathrm{d} \mathbf{L} \mathbf{D} \mathbf{L}^{\top} \right) \\ &= \mathbf{G}^{+} \left(\mathbf{I} + \mathbf{K}_{pp}\right) \left( \mathbf{L} \mathbf{D} \otimes \mathbf{I} \right) \mathrm{vec}\left( \mathrm{d} \mathbf{L} \right) \\ &= 2 \mathbf{G}^{+} \mathbf{G} \mathbf{G}^{+} \left( \mathbf{L} \mathbf{D} \otimes \mathbf{I} \right) \mathbf{F}^{\top} \mathrm{d} \mathrm{vech}\left( \mathbf{L} \right) \\ &= 2 \mathbf{G}^{+} \left( \mathbf{L} \mathbf{D} \otimes \mathbf{I} \right) \mathbf{F}^{\top} \mathrm{d} \mathrm{vech}\left( \mathbf{L} \right), \end{align} $$ and $$ \begin{align} \mathrm{d} \mathrm{vech}\left( \mathbf{\Sigma} \right) &= \mathbf{G}^{+} \mathrm{d} \mathrm{vec}\left( \mathbf{L} \mathbf{L}^{\top} \right) \\ &= \mathbf{G}^{+} \mathrm{vec}\left( \mathrm{d} \mathbf{L} \mathbf{L}^{\top} \right) + \mathbf{G}^{+} \mathrm{vec}\left( \mathbf{L} \mathrm{d} \mathbf{L}^{\top} \right) \\ &= \mathbf{G}^{+} \left(\mathbf{I} + \mathbf{K}_{pp}\right) \mathrm{vec}\left( \mathrm{d} \mathbf{L} \mathbf{L}^{\top} \right) \\ &= \mathbf{G}^{+} \mathbf{G} \mathbf{G}^{+} \mathrm{vec}\left( \mathrm{d} \mathbf{L} \mathbf{L}^{\top} \right) \\ &= 2 \mathbf{G}^{+} \left(\mathbf{L} \otimes \mathbf{I}\right) \mathrm{vec}\left( \mathrm{d} \mathbf{L} \right) \\ &= 2 \mathbf{G}^{+} \left(\mathbf{L} \otimes \mathbf{I}\right) \mathbf{F}^{\top} \mathrm{d} \mathrm{vech}\left( \mathbf{L} \right). \end{align} $$