Here is a concrete counterexample.

Let $X$ be a degree $4$ plane curve with a unique node $r \in X$. The normalization $f \colon \tilde{X} \to X$ is a genus $2$ curve, hence hyperelliptic. Let $p, q \in \tilde{X}$ be two points that are conjugate under the hyperelliptic involution and do not map to $r$ under the normalization map. Considering $p$ and $q$ as points of $X$, I claim that the line bundle $\mathcal{L} := \mathcal{O}_{X}(p+q)$ has the desired properties.

Indeed, we have $h^{0}(X, \mathcal{L})=1$, but $h^{0}(\tilde{X}, g^{*}(\mathcal{L}))=2$. For the second equality, just observe that $g^{*}(\mathcal{L})$ is the line bundle associated to the Cartier divisor $p, q$, now considered as points of $\tilde{X}$. These points are conjugate, so $g^{*}(\mathcal{L})$ is the pullback of $\mathcal{O}(1)$ under the degree $2$ map $\tilde{X} \to \mathbb{P}^1$.

On the other hand, $h^{0}(X, \mathcal{L})=1$. This can be seen using Riemann-Roch. If $\omega$ is the dualizing sheaf, then it is equivalent to show that $h^{0}(X, \omega(-p-q))=1$. But by adjunction, $\omega$ is just the restriction of $\mathcal{O}_{\mathbb{P}^2}(1)$, and certainly there is just a $1$-dimensional space of

Here is a concrete counterexample.

Let $X$ be a degree $4$ plane curve with a unique node $r \in X$. The normalization $f \colon \tilde{X} \to X$ is a genus $2$ curve, hence hyperelliptic. Let $p, q \in \tilde{X}$ be two points that are conjugate under the hyperelliptic involution and do not map to $r$ under the normalization map. Considering $p$ and $q$ as points of $X$, I claim that the line bundle $\mathcal{L} := \mathcal{O}_{X}(p+q)$ has the desired properties. Indeed, I claim that we have $h^{0}(X, \mathcal{L})=1$, but $h^{0}(\tilde{X}, f^{*}(\mathcal{L}))=2$.

For the second equality, just observe that the line bundle $f^{*}(\mathcal{L})$ is just the line bundle associated to a Cartier divisor given by a conjugate pair of points. In other words, this line bundle is the pullback of $\mathcal{O}(1)$ on $\mathbb{P}^1$ under the degree $2$ map $\tilde{X} \to \mathbb{P}^1$. This has a $2$-dimensional space of global sections, coming from the global sections over $\mathbb{P}^1$.

The first equality can be seen using Riemann-Roch. Examining the Riemann-Roch formula, we see this equality is equivalent to the equality $h^{0}(X, \omega(-p-q))=1$, where $\omega$ is the dualizing sheaf. But $\omega$ is just the restriction of $\mathcal{O}(1)$ on $\mathbb{P}^2$ (adjunction!), so the vector space in question is just homogeneous polynomials in $X, Y, Z$ of degree $1$ that vanish at two distinct points. Linear algebra shows that this is a $1$-dimensional space.