Your own reasoning here is flawed because ---although it is true that the respective coefficients appear in the model equation in the same way--- the penalties applying to the fit are different for the two cases. Ultimately, we need to figure out which method penalises analogous coefficients more heavily. I'm going to try to give you an intuitive answer, appealing to a comparison where we first optimise the "split" between the two coefficients in the sum. (I'm also going to assume that you intended for your model equations to contain error terms, so that these are (non-deterministic) statistical models.)
Reframing the problem: Since you have stipulated that $X_1=X_2$, let's simplify your model expressions to:
$$\begin{matrix}
\text{Model } 1 & & & Y = \alpha_1 X_1 + \alpha_3 X_3 + \varepsilon, \quad \quad \quad \ \\[8pt]
\text{Model } 2 & & & Y = (\beta_1+\beta_2) X_1 + \beta_3 X_3 + \varepsilon_*. \\[6pt]
\end{matrix}$$
Note that this is a case where the second model is non-identifiable, because we have split the identifiable parameter $\alpha_1$ into the sum $\beta_1+\beta_2$ containing the non-identifiable parameters $\beta_1$ and $\beta_2$. Under ridge regression we penalise proportionately to the norm of the coefficient vector, so the question here hinges on whether we will reduce or increase this norm (and thus the penalty) by splitting the identifiable parameter into a sum of non-identifiable parameters. To check this, we will first examine the optimal "split" in the sum $\beta_1+\beta_2$ and then compare the optimisations.
Formal analysis: Since the residual-sum-of-squares in Model 2 depends on $\beta_1$ and $\beta_2$ only through their sum, let us first inquire as to how we can form these parameters to minimise their penalty, conditional on a particular sum. This is the constrained optimisation problem:
$$\text{Minimise } \beta_1^2 + \beta_2^2
\quad \quad \text{subject to} \quad \quad
\beta_1 + \beta_2 = \text{const}.$$
Some simple calculus shows that the minimum occurs when $\beta_1 = \beta_2$. Now, using this as a partial requirement for the overall optimisation, consider the following three objective functions for fitting the models:
$$\begin{matrix}
F_1(\alpha_3, \alpha_3) = \text{RSS}_1 + \lambda (\alpha_1^2 + \alpha_3^2), \quad \ \ \\[8pt]
F_2(\beta_1, \beta_3) = \text{RSS}_2 + \lambda (2\beta_1^2 + \beta_3^2), \quad \\[8pt]
F_2^*(\beta_1, \beta_3) = \text{RSS}_2 + \lambda ((2\beta_1)^2 + \beta_3^2). \\[6pt]
\end{matrix}$$
The functions $F_1$ and $F_2$ are the actual objective functions for the models. The function $F_2^*$ is the minimisation problem that would apply if we fit Model 2 but treated $\beta_1 + \beta_2$ as a single coefficient in the penalty function (i.e., as if it were fitted like Model 1). Since $(2\beta_1)^2 = 4 \beta_1^2 \geqslant 2 \beta_1^2$, we can instantly see that the latter objective has a higher penalty for any value $\beta_1 \neq 0$. This means that Model 1 penalises the coefficient on the first term more heavily and so the coefficient will tend to be smaller in magnitude under this model --- i.e., we have:
$$|\hat{\alpha}_1| \leqslant |\hat{\beta}_1 + \hat{\beta}_2|,$$
and this inequality will tend to be strict for cases where the true coefficient is non-zero. (The exact conditions for a strict inequality are complicated, but roughly speaking this will occur when the RSS is continuously differentiable in $\beta_1$, with a maximum that occurs when $\beta_1 \neq 0$. In this case the higher penalty will push the optima for $\beta_1$ downward in magnitude in a continuous way.) Thus, we see that the magnitude of the slope coefficient for $X_1$ will tend to be smaller under Model 1 than under Model 2.
