In this problem $f(x) = g(h(x))$, where $h(x) = ax + b$. I'm going to consider the case where $a$ is a matrix rather than a scalar, because it's useful and no more difficult. You can assume $a$ is a scalar if you'd like.
Let's establish some notation. Recall that if $F:\mathbb R^n \to \mathbb R^m$ is differentiable at $x$, then $F'(x)$ is an $m \times n$ matrix. In the special case where $m = 1$, $F'(x)$ is a $1 \times n$ matrix. I'm going to use the convention that $\nabla F(x) = F'(x)^T$, so $\nabla F(x)$ is a column vector rather than a row vector. Then $G(x) = \nabla F(x)$ is a function from $\mathbb R^n \to \mathbb R^n$, and $\nabla^2 F(x) = G'(x)$, which is an $n \times n$ matrix.
The chain rule tells us that \begin{align} f'(x) &= g'(h(x))h'(x) \\ &= g'(ax + b) a. \end{align} It follows that \begin{align} \nabla f(x) &= a^T g'(ax+b)^T \\ &= a^T \nabla g(ax + b). \end{align} That is our formula for $\nabla f(x)$.
Preparing to use the chain rule again, we can express $\nabla f(x)$ as $\nabla f(x) = w(h(x))$, where $w(x) = a^T \nabla g(x)$. Note that $w'(x) = a^T \nabla^2 g(x)$. Applying the chain rule to $z(x) = \nabla f(x) = w(h(x))$, we see that \begin{align} \nabla^2 f(x) &= w'(h(x))h'(x) \\ &= a^T \nabla^2 g(ax + b) a. \end{align} This is our formula for $\nabla^2 f(x)$.