Well let's see. Say $$\gamma:(-1,1)\to O(n)$$is differentiable and $\gamma(0)=I$; we need to show that $$\gamma'(0)+\gamma'(0)^T=0.$$
We know that $\gamma(t)\gamma(t)^T=I$. Differentiating that gives $$\gamma'(0)\gamma(0)^T+\gamma(0)\gamma'(0)^T=0.$$Oh. We're done, because $\gamma(0)=I$.
First of all, for dimensionality, you should be able to convince yourself that $\dim \mathfrak{s} = \frac{n(n+1)}{2}$. This can be done easily through counting the number of independent parameters in a symmetric matrix. To prove that $S$ is a manifold, one can either show that $S$ is an open subset of $\mathfrak{s}$ or that $S_n \approx GL_n(\mathbb{R})/O(n)$ using the homogeneous space construction theorem. This also proves that $\dim S = \dim GL_n(\mathbb{R}) - \dim O(n) = n^2 - \frac{n(n-1)}{2} = \frac{n(n+1)}{2}$. Therefore they are of the same dimension.
To further show how $S$ and $\mathfrak{s}$ are related, you should recognize that $\mathfrak{s} = T_IS$ since if $\gamma:(-\epsilon, \epsilon)\to GL_n(\mathbb{R})$ with $\gamma(0) = I$ and $\gamma'(0) = X$, then $\eta(t) = \gamma(t)\gamma^T(t)$ defines a smooth curve in $S$. Then, differentiating we find that
\begin{eqnarray*}
\dot{\eta}(0) & = & \dot{\gamma}(0)\gamma^T(0) + \gamma(0)\dot{\gamma}^T(0) \\
& = & X + X^T
\end{eqnarray*}
which is symmetric, proving that $T_IS \subseteq \mathfrak{s}$. Then by dimensionality equality holds.
What's remarkable is that the exponential map $\exp:\mathfrak{s} \to S$ is actually a global diffeomorphism. We can see this actually just through the eigendecomposition of symmetric matrices. If $A \in \mathfrak{s}$ then $A = PDP^T$ where $D$ is diagonal with arbitrary entries, and $P$ is orthogonal. Then basic properties of the Taylor series of $\exp$ show that
$$
\exp(A) \;\; =\;\; P\exp(D)P^T
$$
where $\exp(D)$ will just have the diagonal entries $e^{d_i}$ which will necessarily be positive and therefore force $\exp(A) \in S$. Similarly, the matrix logarithm, defined by the Taylor series
$$
\log(B) \;\; =\;\; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (I - B)^k
$$
serves as an inverse to $\exp$ since the logarithm carries the same computational property: $\log(B) = \log(QEQ^T) = Q\log(E)Q^T$ with the diagonal entries of $\log(E)$ just being $\log(\lambda_i)$ where $\lambda_i$ are the diagonal entries of $E$.
This answer should hopefully give you some insight into the relationship between symmetric matrices and the symmetric positive definite matrices, however in your prompt you seem to be further curious about the pushforward of the exponential. In other words, while $\exp:\mathfrak{s} \to S$ is a diffeomorphism, we still have the pushforward map
$$
d\exp_X: T_X\mathfrak{s} \to T_{\exp(X)}S.
$$
Computing the purshforward is a much more involved computation, however the fact that $\exp$ is a global diffeomorphism should convince you that $d\exp_X$ is an isomorphism between the tangent spaces of the two manifolds.
Best Answer
Define a norm on $\mathrm{Skew}_3$ by $$ \left\|\begin{bmatrix}0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}\right\| \;=\; \sqrt{a^2+b^2+c^2}. $$ Then, for any $A \in \mathrm{Skew}_3$, the norm $\|A\|$ corresponds to the angle of rotation of $\exp(A)$. Specifically, $\exp(A)$ is a rotation by an angle of $\|A\|$ around an axis parallel to the vector $(a,b,c)$.
Note then that if $A\in\mathrm{Skew}_3$ and $\|A\| = 2\pi$, then $\exp(A)$ is equal to the identity matrix. That is, the entire sphere of radius $2\pi$ centered at the origin in $\mathrm{Skew}_3$ maps to the identity matrix under $\exp$. Thus $\exp$ is not a local diffeomorphism (or even a local homeomorphism) at points on this sphere. The same holds true along the sphere of radius $4\pi$, the sphere of radius $6\pi$, and so forth.
By the way, it should be apparent that $\exp$ is a local homeomorphism near zero. In particular, the entire open ball of radius $\pi$ maps homeomorphically into $SO(3)$, and its image consists of all rotations by angles less than $\pi$. However, $\exp$ is not one-to-one on the boundary of this ball, since $\exp(A) = \exp(-A)$ for any $A\in\mathrm{Skew}_3$ with $\|A\| = \pi$.